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I've a hot pice of metal (temperature is $T_1$ Kelvin) and I put that in water (of temperature $T_2$ Kelvin), I also know the volume of the water $m_1$ liters. Then I put the hot metal in the cold water and the temperature of the water will rise to $T_3$ Kelvin and the temperature of the metal will drop to $T_4$ Kelvin

Question: how can I find the specific heat of the metal?

I kind of need to use this formula:

$$c=\frac{Q}{m\cdot\Delta T}$$

I need this for a project that I'm working on

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closed as off-topic by sammy gerbil, Kyle Kanos, John Rennie, Jon Custer, stafusa Nov 2 '17 at 0:02

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Assuming the system loses (or receives) no heat to the environment (assume it's an isolated system), then both water and metal will end up at the same temperature, say $T_3$, given enough time. As per Thermodynamics, heat exchange between metal and water continues until their temperature difference is $0$.

With $m_m$ the mass of metal, $c$ its heat capacity and with $m_w$ the mass of water, $c_w$ its heat capacity, the heat balance then becomes:

$$m_mcT_1+m_wc_wT_2=m_mcT_3+m_wc_wT_3$$

Extract $c$ easily from there:

$$c=\frac{m_wc_wT_3-m_wc_wT_2}{m_mT_1-m_mT_3}$$

Or:

$$c=c_w\frac{m_w}{m_m}\Big(\frac{T_3-T_2}{T_1-T_3}\Big)$$

No heat was lost or gained, only exchanged.

Note that heat capacity of most materials is somewhat temperature dependent, so the value of $c$ obtained this way is an average over the temperature interval $(T_1,T_3)$. If the temperature dependence was known as a function $c(T)$ then that average over $(T_1,T_3)$ would be found as:

$$c=\frac{1}{T_1-T_3}\int_{T_3}^{T_1}c(T)dT$$


A slightly reformulated way of doing this is as follows.

The heat lost by the metal is: $$\Delta H=m_mc(T_1-T_3)$$ Similarly, the heat gained by the water is: $$\Delta H=m_wc_w(T_3-T_2)$$ Since there no heat is lost or gained (isolation assumption), both are identical, so that: $$m_mc(T_1-T_3)=m_wc_w(T_3-T_2)$$ Isolate $c$ and get the same result as above.

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