A block of ice at 0°C with a mass of 2 kg is placed in a room at 25°C. Assuming no heat is lost to the surroundings, what is the total heat absorbed by the ice to completely melt and reach the room temperature? (Specific heat capacity of water = 4186 J/kg·K, latent heat of fusion of ice = 334,000 J/kg)
Key Concept: Heat Transfer, Phase Change
b) 8.77$\times10^5$, ${J}$
[Solution Description]
The process involves two steps:
1. Melting the ice:
$Q_1 = m \times L_f = 2 \, \text{kg} \times 334,000 \, \text{J/kg} = 668,000 \, \text{J}$
2. Heating the melted water to 25°C:
$Q_2 = m \times c \times \Delta T = 2 \, \text{kg} \times 4186 \, \text{J/kg·K} \times (25 - 0) \, \text{K} = 209,300 \, \text{J}$
Total heat absorbed:
$Q_{\text{total}} = Q_1 + Q_2 = 668,000 + 209,300 = 877,300 \, \text{J} \approx 8.77 \times 10^5 \, \text{J}$
Your Answer is correct.
b) 8.77$\times10^5$, ${J}$
[Solution Description]
The process involves two steps:
1. Melting the ice:
$Q_1 = m \times L_f = 2 \, \text{kg} \times 334,000 \, \text{J/kg} = 668,000 \, \text{J}$
2. Heating the melted water to 25°C:
$Q_2 = m \times c \times \Delta T = 2 \, \text{kg} \times 4186 \, \text{J/kg·K} \times (25 - 0) \, \text{K} = 209,300 \, \text{J}$
Total heat absorbed:
$Q_{\text{total}} = Q_1 + Q_2 = 668,000 + 209,300 = 877,300 \, \text{J} \approx 8.77 \times 10^5 \, \text{J}$