TS EAMCET · Physics · Thermal Properties of Matter
The heat required to convert 8 g of ice at a temperature of \(-20^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\) is [Specific heat capacity of ice \(=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\), specific heat capacity of water \(=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\), latent heat of fusion of ice \(=336 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}\) and latent heat of steam \(\left.=2.268 \times 10^6 \mathrm{Jkg}^{-1}\right]\)
- A 5400 cal
- B 5840 cal
- C 5760 cal
- D 5120 cal
Answer & Solution
Correct Answer
(B) 5840 cal
Step-by-step Solution
Detailed explanation
\(Q_{total} = m c_{ice} \Delta T_{ice} + m L_f + m c_{water} \Delta T_{water} + m L_v\) \(Q_{total} = (0.008 \times 2100 \times 20) + (0.008 \times 336 \times 10^3) + (0.008 \times 4200 \times 100) + (0.008 \times 2.268 \times 10^6)\)…
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