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?

Question

Erika

Physics

12 July, 22:07

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?

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Adelyn Orr · Answer 1

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