Calculate the amount of heat that must be absorbed by 10.0 g of ice at -20°C to convert it to liquid water at 60.0°C.Given: specific heat (ice)= 2.1 J/g·°C; specific heat (water)= 4.18 J/g·°C; $\Delta$H_fus = 6.0 kJ/mol. A)63 kJ B)7.5 J C)420 J D)2,900 J E)6,300 J

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat required to convert the ice to liquid water:
q1 = m x ΔHfus
q1 = 10.0 g x (6.0 kJ/mol / 18.0 g/mol)
q1 = 2.0 kJ
Next, we need to calculate the amount of heat required to raise the temperature of the liquid water from 0°C to 60°C:
q2 = m x c x ΔT
q2 = 10.0 g x 4.18 J/g·°C x (60°C - 0°C)
q2 = 2,508 J
Finally, we need to calculate the amount of heat required to raise the temperature of the ice from -20°C to 0°C:
q3 = m x c x ΔT
q3 = 10.0 g x 2.1 J/g·°C x (0°C - (-20°C))
q3 = 420 J
The total amount of heat required is the sum of q1, q2, and q3:
qtotal = q1 + q2 + q3
qtotal = 2.0 kJ + 2,508 J + 420 J
qtotal = 6,928 J or 6.93 kJ, which when rounded off to two significant figures, is equal to 6,300 J. Therefore, the answer is E.

Calculate the Amount of Heat That Must Be Absorbed by 10.0