AP EAMCET · PHYSICS · Thermal Properties of Matter
The amount of heat needed to heat 200 grams of ice at \(-10^{\circ} \mathrm{C}\) to convert it into water at \(30^{\circ} \mathrm{C}\) is
Specific heat capacity of ice \(=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)
Specific heat capacity of water \(=4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)
Latent heat of fusion of ice \(=3.35 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\)
- A \(96316 \mathrm{~J}\)
- B \(67000 \mathrm{~J}\)
- C \(92116 \mathrm{~J}\)
- D \(71200 \mathrm{~T}\)
Answer & Solution
Correct Answer
(A) \(96316 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
We have \(\Delta \mathrm{Q}=\mathrm{mS}_{\mathrm{i}}\left[-10^{\circ}-0^{\circ}\right]-\mathrm{mL}_{\mathrm{f}}+\mathrm{mS}_{\mathrm{w}}\left[0^{\circ}-30^{\circ}\right]\)…
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