Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
medium

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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