A 44.0-g block of ice at -15.0°C is dropped into a calorimeter (of negligible heat capacity)containing 100g. Of water at 5.0°C.When equilibrium is reached,how much of the ice will have melted? The specific heat of ice is 2090 J/kg ∙ K,that of water is 4186 J/kg ∙ K,and the latent heat of fusion of water is 33.5 × 10⁴ J/kg. A)2.1 g B)21 g C)5.2 g D)52 g E)4.4 g

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The correct answer is A because the heat required to melt the ice is equal to the heat lost by the water. The heat lost by the water is:Q = mcΔT = (100 g)(4186 J/kg ∙ K)(5.0°C - 0°C) = 209300 JThe heat required to melt the ice is:Q = mL = (44.0 g)(33.5 × 10^4 J/kg) = 1474000 JTherefore, the mass of ice that will melt is:m = Q/L = 209300 J / 33.5 × 10^4 J/kg = 6.25 gSo, the correct answer is A) 2.1 g.

A 440-G Block of Ice at -15