Question 5.3.3: A 12.00-g block of copper at 12.0 ºC is immersed in a 5.00-g......
A 12.00-g block of copper at 12.0 ºC is immersed in a 5.00-g pool of ethanol with a temperature of 68.0 ºC. When thermal equilibrium is reached, what is the temperature of the copper and ethanol?
You are asked to calculate the temperature of the copper and ethanol when they reach thermal equilibrium.
You are given the mass of copper and ethanol and the initial temperature of copper and ethanol.
Because the magnitude of energy lost by the ethanol is equal to the energy gained by the copper, q_{Cu} + q_{ethanol} = 0.
Use Equation 5.5 to calculate the energy lost and gained upon reaching thermal equilibrium and then calculate the final temperature.
c, specific heat capacity (J/g·°C) = \frac{q,\text{ heat energy absorbed (J)}}{m, \text{mass (g) }\times \Delta T, \text{change in temperature (°C) }} (5.5)
c_{Cu} = 0.385 J/g · ºC c_{ethanol} = 2.44 J/g · ºC
m_{Cu} = 12.00 g m_{ethanol} = 5.00 g
T_{initial}(Cu) = 12.0 ºC T_{initial}(ethanol) = 68.0 ºC
q_{Cu} + q_{ethanol} = 0
[(12.00 g)(0.385 J/g · ºC)(T_{final} − 12.0 ºC)] + [(5.00 g)(2.44 J/g · ºC)(T_{final} − 68.0 ºC)] = 0
(4.62 × T_{final}) − 55.44 + (12.2 × T_{final}) − 829.6 = 0
16.82 × T_{final} = 885.04
T_{final} = 52.6 ºC
Is your answer reasonable? At thermal equilibrium the two substances in contact with one another have the same temperature, and that temperature lies between the two initial temperatures. Whether the final temperature lies closer to the temperature of the initially hotter or cooler object is related to the mass and specific heat of each object. In this case the final temperature is closer to ethanol’s initial temperature because of its significantly higher specific heat (about six times greater than the specific heat of copper).