Question

calculate the amount of heat released when 27.0 g h₂o is cooled from a liquid at 314 k to a solid at 263 k. the melting point of h₂o is 273 k. other useful information about water is listed. (c_{p,liquid}=4.184 j/(gcdot k)) (c_{p,solid}=2.003 j/(gcdot k)) (delta h_{fusion}=40.7 kj/mol) which of these correctly describes how to break this calculation into steps? cool liquid from 314 k to 273 k, freeze liquid at 273 k, and cool solid to 263 k. freeze liquid at 314 k; cool solid to 263 k

Explanation:

Step1: Calculate heat released when cooling liquid water

The formula for heat $q = mc\Delta T$. The mass of water $m = 27.0\ g$, the specific - heat capacity of liquid water $c_{liquid}=4.184\ J/(g\cdot K)$, and the temperature change $\Delta T=314\ K - 273\ K = 41\ K$.
$q_1=27.0\ g\times4.184\ J/(g\cdot K)\times41\ K$
$q_1 = 27.0\times4.184\times41\ J=4639.848\ J$

Step2: Calculate heat released during freezing

First, convert the mass of water to moles. The molar mass of $H_2O$ is $M = 18.015\ g/mol$. The number of moles $n=\frac{27.0\ g}{18.015\ g/mol}\approx1.5\ mol$. The heat of fusion $\Delta H_{fusion}=40.7\ kJ/mol$. But for freezing, the heat released $q_2 = n\times\Delta H_{fusion}$. Since we are going from liquid to solid, $q_2=1.5\ mol\times40.7\ kJ/mol = 61.05\ kJ=61050\ J$

Step3: Calculate heat released when cooling solid water

The specific - heat capacity of solid water $c_{solid}=2.003\ J/(g\cdot K)$, the mass $m = 27.0\ g$, and the temperature change $\Delta T = 273\ K-263\ K = 10\ K$.
$q_3=27.0\ g\times2.003\ J/(g\cdot K)\times10\ K=540.81\ J$

Step4: Calculate total heat released

$q_{total}=q_1 + q_2+q_3$
$q_{total}=4639.848\ J+61050\ J + 540.81\ J=66230.658\ J\approx66.2\ kJ$

The correct way to break the calculation into steps is: Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.

Answer:

Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.