ICSE Class 10 Physics 2015 Solved Paper

Question : 67

Total: 74

A refrigerator converts 100g of water at 20∘C to ice at −10∘C in 35 minutes.
Calculate the average rate of heat extraction in terms of watts.
Given:
Specific heat capacity of ice =2.1Jg−1‌∘C‌−1.
Specific heat capacity of water =4.2Jg−1 ‌∘C−1.
Specific Latent heat of fusion of ice =336J g−1

Solution:

‌‌ Mass of water ‌=100g=0.1‌kg.
‌‌ Temperature of water ‌=20∘C‌. ‌
Amount of heat extracted to convert water from 20∘C to 0∘C.
Q1‌=mC‌water ‌∆t
‌=0.1×4200×(20−0)
‌=8400J
Amount of heat extracted to convert water at 0∘C to 0∘C ice
Q2‌=mL‌ice ‌=0.1×336×1000
‌=33600J
Amount of heat extracted to convert ice at 0∘C to ice at −10∘C.
Q3‌=mC‌ice ‌∆t
‌=0.1×2.1×103×(0−(−10))
‌=2100J
‌ Total Heat ‌(Q)‌=Q1+Q2+Q3
‌=8400+33600+2100
‌=44100J
∴‌‌Pt‌=Q
‌ or ‌‌‌P‌=‌

‌ Power ‌‌=‌

44100

35×60

=‌

44100

2100

‌=21‌Watt

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