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It depends on how much salt you add. Adding salt increases the boiling temperature but decreases the heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, the total time increases; for larger amounts, it decreases.

Using simple chemistry, we can estimate how much longer boiling would take. The elevation in boiling temperature caused by adding the salt can be expressed as $$\Delta T=Kib$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt ($\mathrm{NaCl}$ dissociates in water) and $b$ is the molality, that, is the number of moles of salt per mass of water.

Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, and $T_i$ and $T_f$ are the initial and final temperatures, respectively. The heat capacity of the solution of water + salt as a function of concentration (by mass) can be found here. Note that adding salt lowers the heat capacity.

Let $m_w$ be the mass of the water and $m_s$ be the mass of salt. $C_w \approx 4200\,\mathrm{J\, kg^{-1}\, K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. To boil, the water without salt needs to absorb heat in the amount of $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of water + salt has more mass and needs to reach a higher temperature, so it needs an amount of heat equal to $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$

Assuming a constant heat source, the time required to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for pure water to boil and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$

Now suppose we use $1\,\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \,\mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\,\mathrm{mol}$. The changes in $C$ and $m$ are negligible for such a small amount of salt, so we get $$\frac{t_2}{t_1}=1.01$$ Thus, it takes about $1\%$ more time if you add a small amount of salt.

Say instead we add $0.2\,\mathrm{kg}$ of salt, corresponding to about $3.4\,\mathrm{mol}$. Then according to the source above, the heat capacity of the solution is going to be $C_{sol}\approx 3300\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore, we get

$$\frac{t_2}{t_1}\approx \frac{3300}{4200} \times 1.2 \times \left( 1 + \frac{3.4}{80} \right) \approx 0.98$$

So if we add a substantial amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution of water + salt heats faster.

It depends on how much salt you add. Adding salt increases the boiling temperature but decreases the heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, the total time increases; for larger amounts, it decreases.

Using simple chemistry, we can estimate how much longer boiling would take. The elevation in boiling temperature caused by adding the salt can be expressed as $$\Delta T=Kib$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt ($\mathrm{NaCl}$ dissociates in water) and $b$ is the molality, that is, the number of moles of salt per mass of water.

Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, and $T_i$ and $T_f$ are the initial and final temperatures, respectively. The heat capacity of the solution of water + salt as a function of concentration (by mass) can be found here. Note that adding salt lowers the heat capacity.

Let $m_w$ be the mass of the water and $m_s$ be the mass of salt. $C_w \approx 4200\,\mathrm{J\, kg^{-1}\, K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. To boil, the water without salt needs to absorb heat in the amount of $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of water + salt has more mass and needs to reach a higher temperature, so it needs an amount of heat equal to $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$

Assuming a constant heat source, the time required to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for pure water to boil and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$

Now suppose we use $1\,\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \,\mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\,\mathrm{mol}$. The changes in $C$ and $m$ are negligible for such a small amount of salt, so we get $$\frac{t_2}{t_1}=1.01$$ Thus, it takes about $1\%$ more time if you add a small amount of salt.

Say instead we add $50\,\mathrm{g}$ of salt, corresponding to about $0.86\,\mathrm{mol}$. Then according to the source above, the heat capacity of the solution is going to be $C_{sol}\approx 3900\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore, we get

$$\frac{t_2}{t_1}\approx \frac{3900}{4200} \times 1.05 \times \left( 1 + \frac{0.86}{80} \right) \approx 0.985$$

So if we add a larger amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution of water + salt heats faster.

Note however that these changes are tiny. As pointed out in the comments, the non-linearity of the boiling temperature elevation law might affect the validity of results for larger concentrations. Moreover, hard-to-account-for effects caused by the change in density and viscosity might be relevant.

It depends on how much salt you add: adding salt increases boiling temperature but decreases heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, total time increases, for larger amounts it decreases.

Using some simple chemistry, we can estimate how long more it would take. The elevation in boiling temperature caused by adding the salt can be expressed as: $$\Delta T=Kbi$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt (in water $\mathrm{NaCl}$ dissociates into two) and $b$ is the molality, that is the number of moles of salt per mass of water.

Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, $T_i$ and $T_f$ the initial and final temperature. The heat capacity of the solution of salt+water as a function of concentration (by mass) can be found here. Notice that adding salt lowers the heat capacity.

Let $m_w$ be the mass of the water, $m_s$ be the mass of salt. $C_w \approx 4200 \mathrm{J kg^{-1} K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. In order to boil, the water without salt needs to absorb heat in the amount of: $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of salt + water has more mass, and needs to reach a higher temperature, so it needs an amount of heat equal to: $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$

Assuming a constant heat source, the time it takes to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for water only to boil, and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is: $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$

So now suppose we use $1\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\mathrm{mol}$. The change in $C$ and $m$ is negligible for such a small amount of salt, so we get: $$\frac{t_2}{t_1}=1.01$$ so it takes about $1\%$ more time if you add a small amount of salt.

Say instead we add $0.2\mathrm{kg}$ of salt, corresponding to about $3.4$ moles. Then according to the source above the heat capacity of the solution is going to be $C_{sol}\approx 3300\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore we get:

$$\frac{t_2}{t_1}\approx \frac{3300}{4200} \times 1.2 \times \left( 1 + \frac{3.4}{80} \right) \approx 0.98$$

So if we add a substantial amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution water+salt heats faster.

It depends on how much salt you add. Adding salt increases the boiling temperature but decreases the heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, the total time increases; for larger amounts, it decreases.

Using simple chemistry, we can estimate how much longer boiling would take. The elevation in boiling temperature caused by adding the salt can be expressed as $$\Delta T=Kib$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt ($\mathrm{NaCl}$ dissociates in water) and $b$ is the molality, that, is the number of moles of salt per mass of water.

Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, and $T_i$ and $T_f$ are the initial and final temperatures, respectively. The heat capacity of the solution of water + salt as a function of concentration (by mass) can be found here. Note that adding salt lowers the heat capacity.

Let $m_w$ be the mass of the water and $m_s$ be the mass of salt. $C_w \approx 4200\,\mathrm{J\, kg^{-1}\, K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. To boil, the water without salt needs to absorb heat in the amount of $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of water + salt has more mass and needs to reach a higher temperature, so it needs an amount of heat equal to $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$

Assuming a constant heat source, the time required to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for pure water to boil and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$

Now suppose we use $1\,\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \,\mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\,\mathrm{mol}$. The changes in $C$ and $m$ are negligible for such a small amount of salt, so we get $$\frac{t_2}{t_1}=1.01$$ Thus, it takes about $1\%$ more time if you add a small amount of salt.

Say instead we add $0.2\,\mathrm{kg}$ of salt, corresponding to about $3.4\,\mathrm{mol}$. Then according to the source above, the heat capacity of the solution is going to be $C_{sol}\approx 3300\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore, we get

$$\frac{t_2}{t_1}\approx \frac{3300}{4200} \times 1.2 \times \left( 1 + \frac{3.4}{80} \right) \approx 0.98$$

So if we add a substantial amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution of water + salt heats faster.

It depends on how much salt you add: adding salt increases boiling temperature but decreases heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, total time increases, for larger amounts it decreases.

Using some simple chemistry, we can estimate how long more it would take. The elevation in boiling temperature caused by adding the salt can be expressed as: $$\Delta T=Kbi$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt (in water $\mathrm{NaCl}$ dissociates into two) and $b$ is the molality, that is the number of moles of salt per mass of water.

Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, $T_i$ and $T_f$ the initial and final temperature. The heat capacity of the solution of salt+water as a function of concentration (by mass) can be found here. Notice that adding salt lowers the heat capacity.

Let $m_w$ be the mass of the water, $m_s$ be the mass of salt. $C_w \approx 4200 \mathrm{J kg^{-1} K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. In order to boil, the water without salt needs to absorb heat in the amount of: $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of salt + water has more mass, and needs to reach a higher temperature, so it needs an amount of heat equal to: $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$

Assuming a constant heat source, the time it takes to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for water only to boil, and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is: $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$

So now suppose we use $1\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\mathrm{mol}$. The change in $C$ and $m$ is negligible for such a small amount of salt, so we get: $$\frac{t_2}{t_1}=1.002$$ so it takes about $0.2\%$ more time if you add a small amount of salt.

Say instead we add $0.2\mathrm{kg}$ of salt, corresponding to about $3.4$ moles. Then according to the source above the heat capacity of the solution is going to be $C_{sol}\approx 3300\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore we get:

$$\frac{t_2}{t_1}\approx \frac{3300}{4200} \times 1.2 \times \left( 1 + \frac{3.4}{80} \right) \approx 0.98$$

So if we add a substantial amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution water+salt heats faster.

It depends on how much salt you add: adding salt increases boiling temperature but decreases heat capacity, so it takes less heat for the solution to boil. For small amounts of salt, total time increases, for larger amounts it decreases.

Using some simple chemistry, we can estimate how long more it would take. The elevation in boiling temperature caused by adding the salt can be expressed as: $$\Delta T=Kbi$$ where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt (in water $\mathrm{NaCl}$ dissociates into two) and $b$ is the molality, that is the number of moles of salt per mass of water.

Heat is absorbed according to the law $$Q = M C (T_f-T_i)$$ where $M$ is the mass, $C$ is the heat capacity, $T_i$ and $T_f$ the initial and final temperature. The heat capacity of the solution of salt+water as a function of concentration (by mass) can be found here. Notice that adding salt lowers the heat capacity.

Let $m_w$ be the mass of the water, $m_s$ be the mass of salt. $C_w \approx 4200 \mathrm{J kg^{-1} K^{-1}}$ is the heat capacity of water, and $C_{sol}$ the heat capacity of the solution. In order to boil, the water without salt needs to absorb heat in the amount of: $$Q_1 = m_w C_w (T_0-T_i)$$ The solution of salt + water has more mass, and needs to reach a higher temperature, so it needs an amount of heat equal to: $$Q_2 = (m_w+m_s) C_{sol} (T_0+\Delta T-T_i)$$

Assuming a constant heat source, the time it takes to heat to a certain temperature is proportional to the amount of heat needed. So if $t_1$ is the time it takes for water only to boil, and $t_2$ is the time it takes for water + salt to boil, the ratio of these two is: $$\frac{t_2}{t_1}=\frac{Q_2}{Q_1}=\frac{C_{sol}}{C_w}\left(1 + \frac{m_s}{m_w}\right)\left(1+ \frac{\Delta T}{T_0-T_i}\right)$$

So now suppose we use $1\mathrm{kg}$ of water and start at $T_i=20^\circ \mathrm{C}$. Say we add $10 \mathrm{g}$ of salt; we can use this online calculator to obtain the number of moles, which in this case is $0.17\mathrm{mol}$. The change in $C$ and $m$ is negligible for such a small amount of salt, so we get: $$\frac{t_2}{t_1}=1.01$$ so it takes about $1\%$ more time if you add a small amount of salt.

Say instead we add $0.2\mathrm{kg}$ of salt, corresponding to about $3.4$ moles. Then according to the source above the heat capacity of the solution is going to be $C_{sol}\approx 3300\,\mathrm{J\, kg^{-1}\, K^{-1}}$. Therefore we get:

$$\frac{t_2}{t_1}\approx \frac{3300}{4200} \times 1.2 \times \left( 1 + \frac{3.4}{80} \right) \approx 0.98$$

So if we add a substantial amount of salt, the lowering in heat capacity is sufficient to offset the increase in boiling temperature. The solution water+salt heats faster.