How much heat must be absorbed by ice of mass m = 1000 g at -10 ˚C to take it to liquid state at 15 ˚C?
$c_{ice}$ = 2220 $J k g^{- 1} K^{- 1}$
$c_{liq}$ = 4190 $J k g^{- 1} K^{- 1}$
$L_{F} = 333$ $k J k g^{- 1}$

418 kJ

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300 kJ

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318 kJ

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400 kJ

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Solution

The correct option is A 418 kJ
Heating process is accomplished in three steps:

Step 1: Sensible heat absorbed to raise temperature of ice from -10 ˚C to 0 ˚C:
$Q_{1}$ = $c_{i c e} m (T_{f} - T_{i})$
= (2220 $J k g^{- 1} K^{- 1}$ ) (1 kg) [0 ˚C - (- 10 ˚C) ]
= 22200 J = 22.2 kJ

Step 2: Latent heat absorbed for phase change from ice to water at 0 ˚C:
$Q_{2}$ = $L_{F} m$
= (333 $k J k g^{- 1}$ ) (1 kg) = $333 k J$

Step 3: Sensible heat absorbed to raise temperature of water from 0 ˚C to 15 ˚C:
$Q_{3}$ = $c_{l i q} m (T_{f} - T_{i})$
= (4190 $J k g^{- 1} K^{- 1}$ ) (1 kg) (15 ˚C – 0 ˚C)
=62850 J = 62.85 kJ
Total heat absorbed = $Q_{1} + Q_{2} + Q_{3}$
$= (22.2 + 333 + 62.85) k J$
$= 418.05 k J = 418 k J$

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How much heat must be absorbed by ice of mass m = 1000 g at -10 ˚C to take it to liquid state at 15 ˚C?​ cice = 2220 Jkg−1K−1 ​ cliq = 4190 Jkg−1K−1 LF=333 kJkg−1

How much heat must be absorbed by ice of mass m = 1000 g at -10 ˚C to take it to liquid state at 15 ˚C?
$c_{ice}$ = 2220 $J k g^{- 1} K^{- 1}$
$c_{liq}$ = 4190 $J k g^{- 1} K^{- 1}$
$L_{F} = 333$ $k J k g^{- 1}$