# heat capacity of a hot metal in cold water [closed]

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, Kyle Kanos, John Rennie, Jon Custer, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.

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.