Question

A calorimeter contains 72.0 g of water at 19.2 oC. A 141 g piece of metal is heated to 89.0 oC and dropped into the water. The entire system eventually reaches 25.5 oC . What is the specific heat of the metal?

Answer 1

Answer:

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

$q_{a} = -q_{b}$                          

$m_{a}C_{a}\Delta T_{a} = -m_{b}C_{b}\Delta T_{b}$

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

$m_{a}$: is the water's mass = 72.0 g

$C_{a}$: is the specific heat of water = 4.184 J/(g°C)            

$T_{i_{a}}$: is the initial temperature of the water = 19.2 °C

$T_{f_{a}}$: is the final temperature of the water = 25.5 °C

$m_{b}$: is the metal's mass = 141 g

$C_{b}$: is the specific heat of metal =?

$T_{i_{b}}$: is the initial temperature of the metal = 89.0 °C

$T_{f_{b}}$: is the final temperature of the water = 25.5 °C

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

$72.0 g*4.184 J/(g^{\circ} C)*(25.5 ^{\circ} C - 19.2 ^{\circ} C) = -141 g*C_{b}*(25.5 ^{\circ} C - 89.0 ^{\circ} C)$            

$C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C)$

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!