Question

A piece of copper weighing 300g is heated to 100°c and is then quickly transferred into a copper calorimeter of mass who containing 10g containing 100g of liquid with unknown specific heat capacity. If the final temperature of the mixture is 40°c calculate the specific heat capacity of the liquid (room temperature is 10°c)

(Specific heat capacity of copper is 400jkg/1

Answer 1

Writing down the termal balance equation, obtain:

$$Q_p =Q_c+Q_l$$

which means that the extra termal energy of the piece of copper $Q_p$ was spent to increase the termal energies of calorimeter $Q_c$ and liquid $Q_l$. Substituting the expressions for energies, obtain:

$$c_cm_p(T_1-T_0) = c_lm_l(T_0-T_2) + c_cm_c(T_0-T_2)$$

where $c_c =400\frac{J}{kg\cdot \degree C}$ is the specific heat capacity of copper, $c_l$ is the unknown specific heat capacity of liquid, $m_p = 300g = 0.3kg, m_c = 10g = 0.01kg, m_l = 100g = 0.1kg$ are the masses of the piece, calorimeter and liquid respectively, $T_0 = 40\degree C$ is the final temperature, $T_1 = 100\degree C$ is the initial temperature of the piece, and $T_2 = 10\degree C$ is the initial temperature of the calorimeter and liquid.

Expressing $c_l$, find:

$$c_l = \dfrac{c_cm_p(T_1-T_0) - c_cm_c(T_0-T_2)}{ m_l(T_0-T_2)} = \dfrac{c_cm_p(T_1-T_0)}{m_l(T_0-T_2)}-\dfrac{c_cm_c}{m_l}\\ c_l =\dfrac{400\cdot 0.3\cdot (100-40)}{0.1\cdot (40-10)} - \dfrac{400\cdot 0.01}{0.1} =2360\dfrac{J}{kg\cdot \degree C}$$

Answer. $2360\dfrac{J}{kg\cdot \degree C}$.