Heat capacity of ice: 2090 J kg-1 °C-1 Heat capacity of water (liquid): 4186 J k

Question

Heat capacity of ice: 2090 J kg-1 °C-1
Heat capacity of water (liquid): 4186 J kg-1 °C-1
Latent heat of fusion: 3.36×105 J kg-1
Latent heat of vaporization: 2.26×107 J kg -1

Give your answers to at least three significant figures. Your answers must be accurate to 1%.
How much heat energy is required to evaporate 676 g of water which is initially at a temperature of 73 °C?

A: 6526382.3449 B: 7374812.0497 C: 8333537.6161 D: 9416897.5062 E: 10641094.1821 F: 12024436.4257 G: 13587613.1611 H: 15354002.8720 J

Explanation / Answer

given

mass of water = 676 g

mass of water = 0.676 kg

now

heat reuqired ot bring water from 73 C to 100 C

Q1 = mass x heat capacity x temp change

Q1 = m x c x dT

Q1 = 0.676 x 4186 x ( 100-73)

Q1 = 76402.872

now

heat required to vaporize

Q2 = mass x dHvap

Q2 = 0.676 x 2.26 x 10^7

Q2 = 15277600

so

total heat = Q1 + Q2

total heat = 76402.872 + 15277600

total heat = 15354002.87 J

so

the answer is option H ) 15354002.8729 J

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