JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 \(\left.=3.5 \times 10^{5} J\,kg ^{-1}\right]\)
- A \(90\)
- B \(89\)
- C \(95\)
- D \(100\)
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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