# Question d6e4b

May 26, 2015

You'd need 200 kJ to turn that much ice at $\text{-10"^@"C}$ into water at ${50}^{\circ} \text{C}$.

The constants that you must use are

Heat of fusion of water: $\Delta {H}_{f}$ = $334$ $\text{J/g}$
Specific heat of ice: $c$ = $2.09$ $\text{J}$/$\text{g"^@"C}$;
Specific heat of water: $c$ = $4.18$ $\text{J}$/$\text{g"^@"C}$;

Notice that you must go from solid ice to liquid water, which implies that you must go through one phase change, solid to liquid.

This means that you're going to need to calculate the heat required to go from ice at $\text{-10"^@"C}$ to ice at ${0}^{\circ} \text{C}$, from ice at ${0}^{\circ} \text{C}$ to water at ${0}^{\circ} \text{C}$, and finally from water at ${0}^{\circ} \text{C}$ to water at ${50}^{\circ} \text{C}$.

• Going from ice at $\text{-10"^@"C}$ to ice at ${0}^{\circ} \text{C}$

${q}_{1} = m \cdot {c}_{\text{ice" * DeltaT_"Ice}}$

q_1 = 400cancel("g") * 2.09"J"/(cancel("g") ^@cancel("C")) * (0 -(-10))^@cancel("C") = "8360 J"

• Going from ice at ${0}^{\circ} \text{C}$ to water at ${0}^{\circ} \text{C}$

${q}_{2} = m \cdot \Delta {H}_{\text{f}}$

q_2 = 400cancel("g") * 334"J"/cancel("g") = "133600 J"

• Going from water at ${0}^{\circ} \text{C}$ to water at ${50}^{\circ} \text{C}$

${q}_{3} = m \cdot {c}_{\text{water" * DeltaT_"water}}$

q_3 = 400cancel("g") * 4.18"J"/(cancel("g") ^@cancel("C")) * (50 - 0)^@"C" = "83600J"

The total heat required will be equal to

${q}_{\text{total}} = {q}_{1} + {q}_{2} + {q}_{3}$

${q}_{\text{total" = 8360 + 133600 + 83600 = "225560 J}}$

Expressed in kJ and rounded to one sig fig, the number of sig figs you gave for the mass of the ice, the answer will be

q_"total" = color(green)("+200 kJ")#