# Heat change of mixing substances in different phases

10 g of water at 70C is mixed with 5g of water at 30C Find temperature of mixture in equilibrium

I solve this problem by first converting the water at 70C to 40C and using the excess energy to figure out how much the combined water should rise up.

However my book had another approach to it as follows:

$$Q_{\text{heat given by steam}} = Q_{\text{heat taken by water}}$$

Then,

$$m_1 c_1 | \Delta t_1 | = m_2 c_2 | \Delta t_2|$$

Now what exactly is the idea behind modulus of the temperature changes?

• The two methods are completely equivalent mathematically, and also equivalent to saying that the change in internal energy $\Delta U$ of the combined system is zero (since no net heat is transferred to, nor work done by the combined system). Commented Oct 4, 2020 at 11:44

The temperature change for the cold body is positive while that for the hot body is negative. You could have also done this as - The process involves heat exchange between the participating bodies only and no heat is lost to the surrounding. So, $$m_1s_1(T-T_1) +m_2s_2(T-T_2) = 0$$ Upon rearranging you will arrive at the given result.

• If I understood correctly, they did this just take care of the signs involved (heat flow only from hot to cold)? Commented Oct 4, 2020 at 10:56
• Yes. When heat is lost $∆T$ will be negative , so you have to take care of signs. Commented Oct 4, 2020 at 11:01

The heat energy gained or lost by a mass $$m$$ when it changes temperature by an amount $$\Delta T$$ is proportional to $$m$$ and to $$\Delta T$$. Since the two quantities involved here are both water, we can expect the constant of proportionality to be the same (in your notation, we can assume that $$c_1=c_2$$). So if the $$10$$ g of water becomes colder by $$\Delta_1 T$$ and the $$5$$ g of water becomes warmer by $$\Delta_2 T$$ then conservation of energy tells us that

$$10 \Delta_1 T = 5 \Delta_2 T$$

But both parts must end up at the same temperature, so

$$\Delta_1 T + \Delta_2 T = 70 - 30 = 40$$

I'll let you take it from there.

If you were mixing two different substances then their specific heat capacities would not be the same ($$c_1 \ne c_2$$) and the problem would become more difficult.