Question

The heat required to convert 8 g of ice at a temperature of \( -20^\circ\text{C} \) to steam at \( 100^\circ\text{C} \) is [Specific heat capacity of ice = \( 2100 \, \text{Jkg}^{-1}\text{K}^{-1} \), specific heat capacity of water = \( 4200 \, \text{Jkg}^{-1}\text{K}^{-1} \), latent heat of fusion of ice = \( 336 \times 10^3 \, \text{Jkg}^{-1} \) and latent heat of steam = \( 2.268 \times 10^6 \, \text{Jkg}^{-1} \)]

• A. 5400 cal
• B. 5840 cal
• C. 5760 cal
• D. 5120 cal

Hint:

Break the problem into individual steps for each phase and temperature interval. Remember to convert the final energy from Joules to Calories if the options require it.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to calculate the total heat energy required for a multi-step process involving temperature changes and phase changes: 1. Ice (\(-20^\circ\)C) \(\to\) Ice (\(0^\circ\)C) 2. Ice (\(0^\circ\)C) \(\to\) Water (\(0^\circ\)C) (Fusion) 3. Water (\(0^\circ\)C) \(\to\) Water (\(100^\circ\)C) 4. Water (\(100^\circ\)C) \(\to\) Steam (\(100^\circ\)C) (Vaporization)
Step 3: Detailed Explanation:
Mass \( m = 8 \, \text{g} = 0.008 \, \text{kg} \). 1. Heat to raise ice temp (\(Q_1\)): \[ Q_1 = m c_{ice} \Delta T = 0.008 \times 2100 \times 20 = 336 \, \text{J} \] 2. Heat to melt ice (\(Q_2\)): \[ Q_2 = m L_f = 0.008 \times 336 \times 10^3 = 2688 \, \text{J} \] 3. Heat to raise water temp (\(Q_3\)): \[ Q_3 = m c_{water} \Delta T = 0.008 \times 4200 \times 100 = 3360 \, \text{J} \] 4. Heat to vaporize water (\(Q_4\)): \[ Q_4 = m L_v = 0.008 \times 2.268 \times 10^6 = 18144 \, \text{J} \] Total Heat \( Q_{total} \) in Joules: \[ Q_{total} = 336 + 2688 + 3360 + 18144 = 24528 \, \text{J} \] Convert Joules to calories (using \( 1 \, \text{cal} \approx 4.2 \, \text{J} \)): \[ Q_{cal} = \frac{24528}{4.2} = 5840 \, \text{cal} \]
Step 4: Final Answer:
The heat required is 5840 cal.