Question

Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?

Answer 1

Answer:

The final temperature is 61.65 °C

Explanation:

mass of copper pot $m_{c}$ = 2 kg

temperature of copper pot $T_{c}$ = 20 °C  (the pot will be in thermal equilibrium with the room)

specific heat capacity of copper $C_{c}$= 385 J/kg-°C

The heat content of the copper pot = $m_{c}$$C_{c}$$T_{c}$ = 2 x 385 x 20 = 15400 J

mass of boiling water $m_{w}$ = 200 g = 0.2 kg

temperature of boiling water $T_{w}$ = 100 °C

specific heat capacity of water $C_{w}$ = 4182 J/kg-°C

The heat content of the water = $m_{w}$$C_{w}$$T_{w}$ = 0.2 x 4182 x 100 = 83640 J

The total heat content of the water and copper mix $H_{T}$ = 15400 + 83640 = 99040 J

This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation

$H_{T}$ =   $m_{c}$$C_{c}$$T_{f}$ + $m_{w}$$C_{w}$

where $T_{f}$ is the final temperature of the water and the copper

substituting values, we have

99040 = (2 x 385 x $T_{f}$) + (0.2 x 4182 x

99040 = 770$T_{f}$ + 836.4

99040 = 1606.4$T_{f}$

$T_{f}$ = 99040/1606.4 = 61.65 °C