Question

A 14.28-g block of solid tin at 11.10 °C is immersed in a 22.40-g pool of liquid propylene glycol with a temperature of 64.85 °C. When thermal equilibrium is reached, what is the temperature of the tin and propylene glycol?

Specific heat capacities: tin = 0.213 J/g °C; propylene glycol = 2.50 J/g °C

________ °C

Answer 1

$$Q_1 = -Q_2$$

$$Q_1 = c_1m_1 (T_2 - T_1) = 14.28 g \times 0.213 \frac{J}{g\times ^\circ C}\times (T_2 - 11.10 ^\circ C)$$

$$Q_2 = c_2m_2(T_2 - T_1) = 22.40 g \times 2.50 \frac{J}{g\times ^\circ C} \times (T_2 - 64.85 ^\circ C)$$

$$14.28 g \times 0.213 \frac{J}{g\times ^\circ C}\times (T_2 - 11.10 ^\circ C) = - (22.40 g \times 2.50 \frac{J}{g\times ^\circ C} \times (T_2 - 64.85 ^\circ C))$$

$$T_2 = 62.08 ^\circ C$$