Question

enthalpy and phase changes
273\tfreeze liquid\t$delta h_{fusion}=40.7\frac{kj}{mol}$\t$q_2 = ndelta h_{fus}$
$273
ightarrow263$\tcool solid\t$c_{p,solid}=2.093\frac{j}{gcdot k}$\t$q_3 = cmdelta t$
$314
ightarrow263$\tliq $
ightarrow$ sol\t\t$q_{tot}=q_1 + q_2+q_3$
$q_1=square kj$
$q_2=square kj$
$q_3=square kj$
$q_{tot}=square kj$

Explanation:

Step1: Identify given values

We are given $\Delta H_{fusion}=40.7\frac{kJ}{mol}$, $C_{p,solid}=2.093\frac{J}{g\cdot K}$. But we are missing values for $n$ (number of moles), $m$ (mass) and some initial - state information for calculating $q_1$. Since we can't calculate $q_1$ without more data, assume we are only focusing on $q_2$ and $q_3$ for now (if applicable). For $q_2 = n\Delta H_{fus}$, we need $n$. For $q_3=cm\Delta T$, we need $m$. Let's assume we have 1 mol for $q_2$ and 1 g for $q_3$ (for illustration purposes, actual values are needed in a real - world scenario).

Step2: Calculate $q_2$

If $n = 1mol$ and $\Delta H_{fusion}=40.7\frac{kJ}{mol}$, then $q_2=n\Delta H_{fus}=1mol\times40.7\frac{kJ}{mol}=40.7kJ$

Step3: Calculate $\Delta T$ for $q_3$

For the temperature change from $T_1 = 273K$ to $T_2 = 263K$, $\Delta T=T_2 - T_1=263K - 273K=- 10K$. If $m = 1g$ and $C_{p,solid}=2.093\frac{J}{g\cdot K}$, then $q_3 = C_{p,solid}m\Delta T=2.093\frac{J}{g\cdot K}\times1g\times(-10K)=-20.93J=-0.02093kJ$

Step4: Calculate $q_{tot}$

$q_{tot}=q_1 + q_2+q_3$. Since $q_1$ is unknown, if we assume $q_1 = 0$ for now, $q_{tot}=0 + 40.7kJ-0.02093kJ = 40.67907kJ$

Answer:

Since we made assumptions for illustration:
$q_1$: Unknown (need more data)
$q_2$: $40.7kJ$
$q_3$: $-0.02093kJ$
$q_{tot}$: $40.67907kJ$ (assuming $q_1 = 0$)