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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
  • JEE Mains 2022
  • Medium
  • Numerical

A block of ice of mass \(120\,g\) at temperature \(0^{\circ} C\) is put in \(300\,gm\) of water at \(25^{\circ} C\). The \(xg\) of ice melts as the temperature of the water reaches \(0^{\circ} C\). The value of \(x\) is [Use: Specific heat capacity of water \(=4200\) \(J\,kg ^{-1} K ^{-1}\), Latent heat of ice \(\left.=3.5 \times 10^{5} J\,kg ^{-1}\right]\)

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Answer & Solution

Correct Answer

(A) \(90\)

Step-by-step Solution

Detailed explanation

Energy released by water

\(=0.3 \times 2.5 \times 4200=31500\,J\)

let \(m \  kg\) ice melts

\(m \times 3.5 \times 10^{5}=31.500\)

\(m =\frac{31500 \times 10^{-5}}{3.5}=9000 \times 10^{-5}\)

\(m =0.09\,kg =90\,gm\)

\(x =90\)

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