Question

How much heat (in kj) is required to warm 11.0 g of ice, initially at -13.0 ∘c, to steam at 111.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘c?

Answer 1

The total heat required will be 33.35 KJ.

Step-by-step explanation:

In this question there are two kinds of processes.

1) Heat required for change in temperate = mC∆T

2) Heat required for state change = mass × heat of fusion or latent heat of vaporization

We need some other constants to solve this question.

Heat of fusion of water = 334 J/g

Latent heat of vaporization = 2230 J/g

Specific heat of water = 4.184 J/g0C

Heat required to change temperature from -13 to 0 degree centigrade

Q1 = (11)(2.09)(13) = 298.87 J

Heat required in converting ice to water at 0 degree centigrade

Q2 = (11)(334) = 3674 J

Heat required to change temperature from 0 to 100 degree centigrade

Q3 = (11)(4.184)(100) = 4602.4 J

Heat required in converting water to water vapors at 100 degree centigrade

Q4 = (11)(2230) = 24530 J

Heat required to change temperature from 100 to 111 degree centigrade

Q5 = (11)(2.01)(11) = 243.21 J

Total heat required will be:

Q = Q1 + Q2 + Q3 + Q4 + Q5

Q = 298.87 + 3674 + 4602.4 + 24530 + 243.21 = 33348.48 J

Q = 33.35 KJ

Answer 2

Heating ice from -13°C to 0°C
Q1 = mCΔT = 11*2.09*(0--13) = 11*2.09*13 = 298.87 J

Melting ice at 0°C, with latent heat of fusion being 333 J/g
Q2 = mC = 11*333 = 3663 J

Heating water from 0°C to 100°C, with specific heat of water being 4.184 J/g.°C
Q3 = mCΔT = 11*4.184*100 = 4602.4 J

Vaporizing water at 100°C with latent heat of vaporization being 2230 J/g
Q4 = mC = 11*2230 = 24530 J

Heating steam from 100°C to 111°C
Q5 = mCΔT = 11*2.01*(111-100) = 243.21 J

Total heat required, Q =Q1+Q2+Q3+Q4+Q5 = 298.87+3663+4602.4+24530+243.21 = 33337.48 J ≈ 33.34 kJ