Calculate the amount of heat that must be absorbed by 10.0 g of ice at -20^$\circ$C to convert it to liquid water at 60.0^$\circ$C. Given: specific heat (ice) = 2.1 J/g·^$\circ$C; specific heat (water) = 4.18 J/g·^$\circ$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 heat required to melt the ice:
Q1 = m * ΔHfus
where m = 10.0 g (mass of ice) and ΔHfus = 6.0 kJ/mol (heat of fusion). We first need to convert the mass of ice to moles:
moles of ice = (10.0 g) / (18.02 g/mol) = 0.5557 mol
Q1 = (0.5557 mol) * (6.0 kJ/mol) = 3.33 kJ

Next, we need to calculate the heat required to raise the temperature of liquid water from -20°C to 60°C:
Q2 = m * c * ΔT
where m = 10.0 g (mass of water), c = 4.18 J/g·°C (specific heat of water) and ΔT = 80°C (final temperature - initial temperature). 
Q2 = (10.0 g) * (4.18 J/g·°C) * (80°C) = 3,344 J

Adding Q1 and Q2, we get the total heat required:
Q = Q1 + Q2 = 3.33 kJ + 3,344 J = 6,674 J = 6,670 J (to 3 significant figures)

Therefore, the answer is E) 6,300 J.

Calculate the Amount of Heat That Must Be Absorbed by 10.0