What is the temperature, at thermal equilibrium, when a 2.55 g sample of cream, originally at 3.89 degrees Celsius, is added to 226.8 g of coffee, originally at 98.3 degrees Celsius? Assume the heat capacity of both cream and coffee are 4.0 J/g degree Celsius
To find the final temperature at thermal equilibrium, we can use the principle of conservation of energy. The heat lost by the coffee will equal the heat gained by the cream.
The heat lost by the coffee can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the coffee:
Qcoffee = (226.8 g) * (4.0 J/g°C) * (Tfinal - 98.3°C)
The heat gained by the cream can be calculated using the same formula:
Qcream = (2.55 g) * (4.0 J/g°C) * (Tfinal - 3.89°C)
Since the heat lost by the coffee is equal to the heat gained by the cream, we can set these equations equal to each other:
(226.8 * 4 * (Tfinal - 98.3)) = (2.55 * 4 * (Tfinal - 3.89))
Simplifying:
907.2Tfinal - 8974.16 = 10.2Tfinal - 39.732
8974.16 - 39.732 = 907.2Tfinal - 10.2Tfinal
8934.428 = 897.0Tfinal
Tfinal = 8934.428 / 897.0 = 9.95°C
Therefore, the final temperature at thermal equilibrium is approximately 9.95 degrees Celsius.