# Question 42dcd

Mar 26, 2017

$7.80 \times {10}^{3} \text{cal}$, which is $7.80 \text{Cal, kcal, food calories,nutritional calories}$, whatever

#### Explanation:

Recall that $q = m C \Delta T$ within a phase and that during melting, $q = m {H}_{f}$

Given that ${C}_{\text{water"=4.18"J/g°C}}$, ${C}_{\text{ice"=2.09"J/g°C}}$, H_(f"water")=334"J/g", and the melting point of water is $0 \text{°C}$

First we need to make the $55.00 \text{g}$ of ice melt, that is raise it from $\text{-"10.0"°C}$ to $0 \text{°C}$ so $\Delta T = 10.0 \text{°C}$.

${q}_{1} = \left(55.00 \text{g")(2.09"J/g°C")(10.0"°C}\right)$, therefore ${q}_{1} = 1 \text{,"149.5"J"~~1","150"J}$

Then we need to actually melt the ice.

${q}_{2} = \left(55.00 \text{g")(334"J/g}\right)$, therefore ${q}_{2} = 18 \text{,"370"J"~~18","370"J}$

Then we need to raise the water to the desired temperature of $57.0 \text{°C}$ from the current $0 \text{°C}$ so $\Delta T = 57.0 \text{°C}$.

${q}_{3} = \left(55.00 \text{g")(4.18"J/g°C")(57.0"°C}\right)$, therefore ${q}_{3} = 13 \text{,"104.3"J"~~13","100"J}$

To find the total heat, we use $q = {q}_{1} + {q}_{2} + {q}_{3}$

q=(1","150"J")+(18","370"J")+(13","100"J")=32","620"J"~~32","600"J"#

Recall that $1 \text{cal"=4.18"J}$, so $\left(1 \text{cal")/(4.18"J}\right) = 1$

$32 \text{,"600"J"xx(1"cal")/(4.18"J")=7799.043"cal"~~7.80xx10^3"cal}$

Note: I used the $\approx$ symbol to show rounding for significant figures.