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Q.20 gm ice at 0°C is mixed with 40 gm water at 10°C. Then the resultant temperature will be (A) 0°C
(B) +5°C
(C) +6.6°C
(D) -5°C

Bihar BsebBihar Board Intermediate 1st Year 2025MCQ· 1mImportance★★★★★
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The heat available from the warm water is not enough to melt all the ice, so the resultant temperature stays at 0°C.

Heat that could be released by 40 g of water cooling from 10°C to 0°C:

Qreleased=mcΔT=40×1×10=400 calQ_{released} = mc\Delta T = 40\times1\times10 = 400\text{ cal}

Heat required to melt all 20 g of ice at 0°C (latent heat of fusion of ice Lf≈80L_f \approx 80 cal/g):

Qneeded=mLf=20×80=1600 calQ_{needed} = mL_f = 20\times80 = 1600\text{ cal}

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