Question

if 0.03 kg of coffee at 89 degrees celsius is poured into a 0.40 kg cup at 24 degree celsius, what is the final temperature of the coffee?

Answer 1

Wrighting down the heat balance equation, obtain:

$$c_{cof}m_{cof}(t_1 - t_0) = c_{cup}m_{cup}(t_0-t_2)$$

where $m_{cof} = 0.03kg, m_{cup} = 0.4kg$ are the masses of the portion coffee and the cup respectively, $t_1 = 89\degree C, t_2 = 24\degree C$ are the initial temperatures of the portion coffee and the cup respectively, and $c_{cof},c_{cup}$ are the specific heat capacities of the portion coffee and the cup respectively. $t_0$ is the final temperature of the system.

Let's assume $c_{cof} = 4200\dfrac{J}{kg\cdot \degree C}$ to be the same as for water and the cup to be made of porcelain with $c_{cup}= 1085\dfrac{J}{kg\cdot \degree C}$. Then expressing $t_0$, obtain:

$$t_0 = \dfrac{c_{cof}m_{cof}t_1 + c_{cup}m_{cup}t_2}{c_{cof}m_{cof} + c_{cup}m_{cup}}\\ t_0 = \dfrac{4200\cdot 0.03\cdot 89 + 1085\cdot 0.4\cdot 24}{4200\cdot 0.03+ 1085\cdot 0.4} =38.625\degree C$$

Answer. $38.625\degree C$.