JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 \(=2.25 \times 10^6\) \(\mathrm{Jkg}^{-1}\) )
- A 3043 J
- B 3024 J
- C 3003 J
- D 3022 J
Answer & Solution
Correct Answer
(A) 3043 J
Step-by-step Solution
Detailed explanation
\begin{aligned} & \Delta \mathrm{Q}_1=\mathrm{m} \times \mathrm{S}_1 \times \Delta \mathrm{T}=10^{-3} \times 2100 \times 10=21 \mathrm{~J} \\ & \Delta \mathrm{Q}_2=\mathrm{m} \times \mathrm{L}_{\mathrm{f}}=10^{-3} \times 3.35 \times 10^5=335 \mathrm{~J} \\ & \Delta…
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