Answer to Question #220587 in Physics for Muhammad

Question

In an attempt to determine the specific heating capacity of copper, the following experiments were used. A copper cylinder weighing 145 g is placed in a large tub of water with a temperature of 94.0  oC. When the cylinder has been in this warm water for along time, it is lifted up and put in a thermos with 335 g of water with a temperature of 12.0  oC. This water is then heated up to 15.4  oC. What value of copper's specific heat capacity does this experiment provide? The specific heat capacity for water is 4.18·1

Answer 1

The equaion of heat balance in this case:

$$c_cm_c(T_{ic} - T_0) = c_wm_w(T_0-T_{iw})$$

where $c_c, \space c_w = 4.18\times 10^3\dfrac{J}{kg\cdot \degree C}$ are specific heat capacities of cooper and water respectively, $m_c = 145g = 0.145kg, \space m_w = 0.335kg$ are the masses of the cooper and water respectively, $T_{ic} = 94\degree C, \space T_{iw} = 12\degree C$ are initial temperatures of the cooper and water respectively, and $T_0 = 15.4\degree C$ is the equilibrium temperature.

Thus, expressing $c_c$, obtain:

$$c_c = \dfrac{c_wm_w(T_0-T_{iw})}{m_c(T_{ic} - T_0)}\\ c_c = \dfrac{4.18\times 10^3\cdot 0.335\cdot (15.4-12)}{0.145\cdot (94 - 15)} \approx 416\dfrac{J}{kg\cdot \degree C}$$

Answer. $416\dfrac{J}{kg\cdot \degree C}$.