Question

24.3 grams of a metal was heated to 95.0°C and placed immediately into an insulated container containing 50.0 grams of water at 25.0°C. The final temperature of the water is 26.8 °C. What is the specific heat of the metal?

Answer 1

As the container is insulated, the quantity of heat that metal lost is equal to the quantity of heat received by water:

$Q_m = -Q_w$

The quantity of heat transferred to water is related to its change in temperature as follows:

$Q_w = c_wm_w(T_{fin} - T_{init, w})$ ,

where $c_w$ is the specific heat of water, $m_w$ is the mass of water and $(T_{fin} - T_{init, w})$ is the change in temperature. We can write an analogical expression for the metal:

$Q_m = c_mm_m(T_{fin} - T_{init, m})$

Let's join the last two equations and write the expression for the specific heat of the metal:

$c_wm_w(T_{fin} - T_{init, w}) = - c_mm_m(T_{fin} - T_{init, m})$

$c_m = -\frac{c_wm_w(T_{fin} - T_{init, w})}{m_m(T_{fin} - T_{init, m})}$

The specific heat of water is 4.186 J/g/°C. As you can see, we don't need to convert the temperature in kelvin here, as we use difference temperature. Finally, the specific heat of the metal is:

$c_m = -\frac{4.186\cdot 50.0\cdot (26.8 - 25.0)}{24.3\cdot(26.8 - 95.0)} = 0.227$ J/g/°C

Answer: The specific heat of the metal is 0.227 J/g/°C. With a big probability, this metal was tin.