Using the values for the heat of fusion, specific heat of water, and/or heat of

Question

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following: joules needed to melt 50.0 g of ice at 0 degree C and to warm the liquid to 65.0 degree C, kilocalories released when 15.0 g of steam condenses at 100 degree C and the liquid cools to 0 degree C kilojoules needed to melt 24.0 g of tee at 0 degree C, w arm the liquid to 100 degree C, and change it to steam at 100 degree C The following graph is a heating curve for chloroform, a solvent for fats, oils, and waxes: (3.7)

Explanation / Answer

Latent heat of fusion of water, Hf = 333.55 J/g = 79.72 cal/g

Latent heat of vaporization of water, Hv = 2257 J/g = 539.44 cal/g

Specific heat capacity of water, Cpw = 1 cal/ g C = 4.187 J/g C

Specific heat capacity of ice, Cpi = 0.504 cal/g C = 2.108 J/ g C

Specific heat capacity of steam, Cpa = 0.477 cal/g C = 1.996 J/ g C

a)

Mass, M = 50 g

T = (65 – 0) C

Heat required = Heat of fusion + Liquid sensible heat

= M * Hf + M * Cpw * T

= 50 * 333.55 + 50 * 4.187 * 65

= 30285.25 J

b)

Mass, M = 15 g

T = (100 – 0) C

Heat released = Heat of vaporization + Liquid sensible heat

= M * Hv + M * Cpw * T

= 15 * 539.44 + 15 * 1 * 100 = 9592.62 cal

= 9.59 kcal

c)

Mass, M = 24 g

T = (100 – 0) C

Heat required = Heat of fusion + Liquid sensible heat + Heat of vaporization

= M * Hf + M * Cpw * T + M * Hv

= 24 * 333.55 + 24 * 4.187 * 100 + 24 * 2257 = 72222 J

= 72.2 kJ

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