Question

An amount of ice of mass \( 10^{-3} \) kg and temperature \( -10^\circ C \) is transformed to vapor of temperature \( 110^\circ C \) by applying heat. The total amount of work required for this conversion is,
(Take, specific heat of ice = 2100 J kg\(^{-1}\) K\(^{-1}\),
specific heat of water = 4180 J kg\(^{-1}\) K\(^{-1}\),
specific heat of steam = 1920 J kg\(^{-1}\) K\(^{-1}\),
Latent heat of ice = \( 3.35 \times 10^5 \) J kg\(^{-1}\),
Latent heat of steam = \( 2.25 \times 10^6 \) J kg\(^{-1}\))

• A. 3022 J
• B. 3043 J
• C. 3003 J
• D. 3024 J

Hint:

To calculate the total heat required, add up the heat for each phase change and temperature change step.

Answer 1

The Correct Option is B

To find the total work done, we calculate the heat required at each step. \[ \Delta Q_1 = m S_i \Delta T = 10^{-3} \times 2100 \times 20 = 21 \, \text{J} \] \[ \Delta Q_2 = m L_i = 10^{-3} \times 3.35 \times 10^5 = 335 \, \text{J} \] \[ \Delta Q_3 = m S_w \Delta T = 10^{-3} \times 4180 \times 100 = 418 \, \text{J} \] \[ \Delta Q_4 = m L_w = 10^{-3} \times 2.25 \times 10^6 = 2250 \, \text{J} \] \[ \Delta Q_5 = m S_s \Delta T = 10^{-3} \times 1920 \times 110 = 19.2 \, \text{J} \] Thus, the total work required is: \[ \Delta Q_{\text{total}} = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \, \text{J} \] Thus, the answer is \( \boxed{3043} \, \text{J} \).