Thermal Properties of Matter

JEE Main level, numerical answer, medium

61 of 126

Question 61 of 126

A block of ice of mass 120g120g at temperature 0∘C0^{\circ}C is put in 300g300g of water at 25∘C25^{\circ}C. The xgxg of ice melts as the temperature of the water reaches 0∘C0^{\circ}C. The value of xx is
[Use: Specific heat capacity of water =4200Jkg−1K−1=4200J{kg}^{-1}{K}^{-1}, Latent heat of ice =3.5×105Jkg−1=3.5\times {10}^{5}J{kg}^{-1}]

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Answer: 90

According to the principle of calorimetry, heat lost by hot body is equal to the heat gained by cold body.
The heat released by water
∆Q=ms∆T=0.3×4200×25=31500J∆Q=ms∆T=0.3\times 4200\times 25=31500J
let mkgmkg ice melts then,
∆Q=mL=m×3.5×105=31500J∆Q=mL=m\times 3.5\times {10}^{5}=31500J
⇒m=31500×10−53.5=9000×10−5\Rightarrow m=\dfrac{31500\times {10}^{-5}}{3.5}=9000\times {10}^{-5}
m=0.09kg=90gm=0.09kg=90g
⇒x=90\Rightarrow x=90

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