JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 9)
An amount of ice of mass $10^{-3} \mathrm{~kg}$ and temperature $-10^{\circ} \mathrm{C}$ is transformed to vapour of temperature $110^{\circ} \mathrm{C}$ by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice $=2100 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, specific heat of water $=4180 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, specific heat of steam $=1920 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, Latent heat of ice $=3.35 \times 10^5 \mathrm{Jkg}^{-1}$ and Latent heat of steam $=2.25 \times 10^6$ $\mathrm{Jkg}^{-1}$ )
3022 J
3043 J
3003 J
3024 J
Explanation
_22nd_January_Morning_Shift_en_9_1.png)
$$\Delta {Q_1} = m{s_i}\Delta {T_2},\,\Delta {Q_2} = m{L_i}$$
$$\Delta {Q_3} = m{S_w}\Delta {T_3},\,\Delta {Q_4} = m{L_s},\,\Delta {Q_5} = m{S_s}\Delta {T_5}$$
$$\Delta {Q_{net}} = m({S_i}\Delta {T_1} + {L_i} + {S_w}\Delta {T_3} + {L_s} + {S_s}\Delta {T_s})$$
$$ = {10^{ - 3}}(2100 \times 10 + 335000 + 4180 \times 100 + 2250000 + 1920 \times 10)$$
$$ = {10^{ - 3}}(3043200) = 3043.2\,J$$
So, $$\Delta {Q_{net}} \approx 3043\,J$$
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