Question

A piece of copper (12.0 g) is heated to 100.0 °C. A piece of chromium (also 12.0 g) is chilled in an ice bath to 0 °C. The specific heat capacity of water is 4.184 J/g ⋅°C.
Both pieces of metal are placed in a beaker containing 200.0 g at 20.0 °C.
(a) Will the temperature of the water be greater than or less than 20.0 °C when thermal equilibrium is reached?
(b) Both pieces of metal are placed in a beaker containing 200.0 g at 20.0 °C. Calculate the final temperature.

Answer 1

Answer:

(a) Slightly greater than 20.0 °C

(b) $T_F=293.46K=20.3^oC$

Explanation:

Hello,

In this case, since we are talking about the equilibrium temperature that will be reached when the copper, chromium and water samples get in contact, the following equation is useful to describe such situation:

$\Delta H_{water}+\Delta H_{copper}+\Delta H_{chromium}=0$

Thus, in terms of masses, heat capacities and temperatures we consider the final temperature as the unknown:

$m_{water}Cp_{water}(T_F-T_{water})+m_{copper}Cp_{copper}(T_F-T_{copper})+m_{chromium}Cp_{chromium}(T_F-T_{chromium})=0$In such a way, by knowing that the heat capacities of copper and chromium are 0.386 and 0.45 J/(g°C) respectively, by solving for the equilibrium temperature one has:

$T_F=\frac{m_{water}Cp_{water}T_{water}+m_{Cu}Cp_{Cu}T_{Cu}+m_{Cr}Cp_{Cr}T_{Cr}}{m_{water}Cp_{water}+m_{Cu}Cp_{Cu}+m_{Cr}Cp_{Cr}}$

$T_F=\frac{200.0g*4.184\frac{J}{g*K}* 293.15K+12.0g*0.386\frac{J}{g*K} *373.15K+12.0g*0.45\frac{J}{g*K}*273.15K}{200.0g*4.184\frac{J}{g*K}+12.0g*0.386\frac{J}{g*K} +12.0g*0.45\frac{J}{g*K}}\\\\T_F=293.46K=20.3^oC$

Hence, the resulting temperature of water turns out slightly greater than 20.0 °C.

Best regards.