# Question 2bf52

Apr 18, 2015

Since your question lacks the starting temperature of the ice, I'll pick one myself. So, I assumed that the initial temperature of the ice is $- {20}^{\circ} \text{C}$.

You have to calculate the energy needed to go from ice at $- {20}^{\circ} \text{C}$ to ice at ${0}^{\circ} \text{C}$, then to go from ice at ${0}^{\circ} \text{C}$ to water at ${0}^{\circ} \text{C}$, then finally from water at ${0}^{\circ} \text{C}$ to water at ${10}^{\circ} \text{C}$.

There are a couple of constants you must use in order to be able to solve this problem

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: $c$ = $4.18$ $\text{J}$/$\text{g"^@"C}$;

So, in order, the steps you're going to go through are

• Calculate the heat required to go from ice at $- {20}^{\circ} \text{C}$ to ice at ${0}^{\circ} \text{C}$

${q}_{1} = m \cdot {c}_{\text{ice}} \cdot \Delta {T}_{1}$, where

$m$ - the mass of the ice;
$\Delta {T}_{1}$ - the change in temperature, defined as ${T}_{\text{final}}$ minus ${T}_{\text{initial}}$;
${c}_{\text{ice}}$ - the specific heat of ice;

${q}_{1} = 20 \cancel{\text{g") * 2.09"J"/(cancel("g") * ^@cancel("C")) * (0 - (-20))^@cancel("C}}$

${q}_{1} = \text{836 J}$

• Calculate the heat required to go from ice at ${0}^{\circ} \text{C}$ to water at ${0}^{\circ} \text{C}$

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

$\Delta {H}_{f}$ - the heat of fusion of water;

q_2 = 20cancel("g") * 334"J"/cancel("g") = "6680 J"

• Calculate the heat required to go from water at ${0}^{\circ} \text{C}$ to water at ${10}^{\circ} \text{C}$

${q}_{3} = m \cdot {c}_{\text{water}} \cdot \Delta {T}_{3}$

${q}_{3} = 20 \cancel{\text{g") * 4.18"J"/(cancel("g") * ^@cancel("C")) * (10 - 0)^@cancel("C}}$

${q}_{3} = \text{836 J}$

The total heat required will be

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

${q}_{\text{total" = "836 J" + "6680 J" + "836 J" = "+8352 J}}$

Expressed in kJ and rounded to one sig fig, the answer will be

q_"total" = color(green)("+8 kJ")#