Question

The amount of heat needed to heat \( 200 \, \text{g} \) of ice at \(-10^\circ C\) to convert it into water at \(30^\circ C\) is: Specific heat capacity of ice = \( 2100 \, \text{J/kgK} \)
Specific heat capacity of water = \( 4186 \, \text{J/kgK} \)
Latent heat of fusion of ice = \( 3.35 \times 10^5 \, \text{J/kg} \)

• A. \( 96316 \, \text{J} \)
• B. \( 67000 \, \text{J} \)
• C. \( 92116 \, \text{J} \)
• D. \( 71200 \, \text{J} \)

Hint:

Always split the heat calculation into three parts: heating ice to \(0^\circ C\), melting it, and then heating water to the final temperature.

Answer 1

The Correct Option is A

Convert mass: \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) 1. Heating ice from \(-10^\circ C\) to \(0^\circ C\): \[ Q_1 = mc\Delta T = 0.2 \times 2100 \times 10 = 4200 \, \text{J} \] 2. Melting ice at \(0^\circ C\): \[ Q_2 = mL = 0.2 \times 3.35 \times 10^5 = 67000 \, \text{J} \] 3. Heating water from \(0^\circ C\) to \(30^\circ C\): \[ Q_3 = mc\Delta T = 0.2 \times 4186 \times 30 = 25116 \, \text{J} \] Total heat: \[ Q = Q_1 + Q_2 + Q_3 = 4200 + 67000 + 25116 = 96316 \, \text{J} \]