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10-1.Thermometry, Thermal Expansion and Calorimetry
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A $2\,kg$ copper block is heated to $500^o\,C$ and then it is placed on a large block of ice at $0^o\,C$. If the specific heat capacity of copper is $400\, J/kg/ ^o\,C$ and latent heat of fusion of water is $3.5 \times 10^5\, J/kg$, the amount of ice, that can melt is :-
A
$(7/8)\, kg$
B
$(7/5)\, kg$
C
$(8/7)\, kg$
D
$(5/7)\, kg$
Solution
Heat gained by ice $=$ heat lost by copper
$m \times 3.5 \times 10^{5}=2 \times 400 \times(500-0)$
where, $m$ is the mass of ice melts
Thus, $\mathrm{m}=\frac{2 \times 400 \times 500}{3.5 \times 10^{5}}=\frac{8}{7} \mathrm{kg}$
Standard 11
Physics
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