Question #94064
How much energy (in kilojoules) is needed to heat 5.40 g of ice from -14.0 ∘C to 31.0 ∘C? The heat of fusion of water is 6.01kJ/mol, and the molar heat capacity is 36.6 J/(K⋅mol) for ice and 75.3 J/(K⋅mol) for liquid water.
1
Expert's answer
2019-09-09T04:59:46-0400

Number of moles of 5.4gm5.4 gm ice=5.418=0.3=\frac{5.4}{18}=0.3

Heat required to raise temperature from 14°-14\degree to zero=ncδt=0.3×=nc \delta t=0.3 \times 36.6×14×103=153.72×103J36.6 \times 14 \times 10^{-3}=153.72 \times 10^{-3} J

Heat required for fusion=6.01×103×0.3=1.803×103J=6.01 \times 10^{3} \times 0.3=1.803 \times 10^3J

Heat required to raise temperature from zero to 31°C=31×75.3×103×0.3=700.29×10331 \degree C=31 \times 75.3 \times 10^{-3} \times 0.3=700.29 \times 10^{-3}

Total heat required=1.803×103+153.72×103+700.29×103==1.803 \times 10^3+153.72 \times 10^{-3}+700.29 \times 10^{-3}= 1803+0.854=1803.854J1804J1803+0.854=1803.854 J\approx 1804 J


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