A 2294-kg sample of water at 0° C is cooled to $- 36 ^ { \circ } \mathrm { C }$ and freezes in the process.How much thermal energy is released? For water L_F = 334,000 J/kg and L_V= 2.256 × 10⁶ J/kg.The specific heat of ice is 2050 J/kg ∙ K.

Question

Quizplus · Accepted Answer

The thermal energy released includes the energy released during freezing and the energy released when cooling the ice. First, calculate the energy released during freezing: $ Q_1 = m \times L_F = 2294 \, \text{kg} \times 334,000 \, \text{J/kg} = 766,196,000 \, \text{J} $. Then, calculate the energy released when cooling the ice from 0°C to -36°C: $ Q_2 = m \times c \times \Delta T = 2294 \, \text{kg} \times 2050 \, \text{J/kg} \cdot \text{K} \times 36 \, \text{K} = 169,452,600 \, \text{J} $. The total energy released is $ Q_1 + Q_2 = 935,648,600 \, \text{J} \approx 935,000 \, \text{kJ} $.

A 2294-Kg Sample of Water at 0° C Is Cooled −36∘C- 36 ^ { \circ } \mathrm { C }−36∘C

A 2294-Kg Sample of Water at 0° C Is Cooled $- 36 ^ { \circ } \mathrm { C }$