Question

Consider the specific heats

H2O(s) = 2.09 J/g ·
◦C,
H2O (ℓ) = 4.18 J/g ·
◦C, and
H2O(g) = 2.03 J/g ·
◦C.
The heat of fusion for water is 334 J/g and
its heat of vaporization is 2260 J/g. Calculate
the amount of heat required to convert 31 g
of ice at −21◦C completely to liquid water at
42◦C.
Answer in units of kJ.

Answer 1

First, we need to calculate the amount of heat required to raise the temperature of ice from -21°C to 0°C:

q1 = m × C_solid × ΔT = 31 g × 2.09 J/g·°C × (0°C - (-21°C)) = 1341.09 J

Next, we need to calculate the amount of heat required to melt the ice at 0°C:

q2 = m × ΔH_fusion = 31 g × 334 J/g = 10354 J

Then, we need to calculate the amount of heat required to raise the temperature of the liquid water from 0°C to 42°C:

q3 = m × C_liquid × ΔT = 31 g × 4.18 J/g·°C × (42°C - 0°C) = 5201.56 J

Finally, we need to calculate the amount of heat required to vaporize the liquid water at 100°C:

q4 = m × ΔH_vaporization = 31 g × 2260 J/g = 70060 J

The total amount of heat required is the sum of all four steps:

q_total = q1 + q2 + q3 + q4 = 86857.65 J

Converting to kJ:

q_total = 86.85765 kJ

Therefore, the amount of heat required to convert 31 g of ice at -21°C to liquid water at 42°C is approximately 86.9 kJ.

Step-by-step explanation: