How much heat is required to raise the temperature of 27.8 g27.8 g of water from −55 °C−55 °C to 145 °C?
Answer
89.4 kJ
Explanation
i) Heat required to raise the temperature of ice from -55°C to 0℃(q1)
q1 = m × ∆T × Cs
= 27.8g × 55°C × 2.108J/g℃
= 3223.1J
ii) Heat required to melt ice at 0℃ (q2)
q2 = m × ∆Hvap
= 27.8g × 333.6J/g
= 9274.1J
iii) Heat required to raise the temperature of liquid water from 0℃ to 100℃ (q3)
q3 = m × ∆T × Cs
= 27.8g × 100℃ × 4.184J/g℃
= 11632J
iv) Heat required to vaporise liquid water at 100℃ (q4)
q4 = m × ∆Hvap
= 27.8 × 2257J/g
= 62745J
v) Heat required to raise the temperature of gaseous water from 100℃ to 145℃(q5)
q5 = m ×∆T × Cs
= 27.8g × 45℃ × 1.996J/g℃
= 2497J
vi) Total heat required(q)
q = q1 + q2 + q3+ q4 + q5
= 3223.1J + 9274.1J + 11632J + 62745J + 2497J
= 89371J
= 89.4kJ
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