10 g of ice at 0∘C absorbs 5460 J of heat energy to melt and change to water at 50∘C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200Jkg−1K−1.
2168Jg−1
336Jg−1
21.68Jg−1
43.36Jg−1
A
2168Jg−1
B
21.68Jg−1
C
336Jg−1
D
43.36Jg−1
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Solution
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Heat absorb =mL+mc×(T−0)
5460=10×L+10×4.2×50
L=336Jg−1
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