Thermal Properties of Matter

JEE Main level, single correct, medium

33 of 126

Question 33 of 126

The specific heat of water =4200J=4200\text{J} kg–1K–1{\text{kg}}^{–1}{\text{K}}^{–1} and the latent heat of ice =3.4×105J=3.4\times {10}^{5}\text{J} kg–1{\text{g}}^{–1}. 100100 grams of ice at 0oC0{ }^{\text{o}}\text{C} is placed in 200g200\text{g} of water at 25oC25{ }^{\text{o}}\text{C}. The amount of ice that will melt as the temperature of water reaches 0oC0{ }^{\text{o}}\text{C} is close to (in grams)

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Answer: (A) 61.761.7

MSΔT=MlMS\Delta T=Ml
2001000×4200×25=m×340×103\dfrac{200}{1000}\times 4200\times 25=m\times 340\times {10}^{3}
m=61.7m=61.7

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