## Related questions with answers

What $\mathrm{Btu}$ of heat are released when $20.0 \mathrm{~lb}$ of water at $80{\degree} \mathrm{F}$ is cooled to $32{\degree} \mathrm{F}$ and then frozen in an ice plant?

Solution

Verified$\textbf{Given:}$

$w=20 \text{ lb}$

$T_1=32 ^\circ\text{F}$

$T_2=80 ^\circ\text{F}$

$L_f=144 \text{ } \dfrac{ \text{Btu}}{ \text{ lb}}$

$c=1 \text{ }\dfrac{ \text{Btu}}{ \text{lb}^\circ\text{F}}$

We first calculate the heat required to cool water to $32 ^ \circ \text {F}$ using the heat formula:

$Q_1=w\cdot c \cdot \Delta T$

By entering the given data we get:

$\begin{align*} Q_1&=w\cdot c \cdot (T_2-T_1)\\ &=20 \cdot 1 \ (80-32)\\ &\boxed{Q_1=960 \text{ Btu}} \end{align*}$

Now let's calculatethe heat required to freez the water To do this we use the definition of heat of fusion given by:

$L_f=\dfrac{Q_2}{w}$

We express the heat and include the given data:

$\begin{align*} Q_2&=L_f \cdot w\\ &=144 \cdot 20\\ &\boxed{Q_2=2880 \text{ Btu}} \end{align*}$

To get an overall heat add up both results:

$\begin{align*} Q&=Q_1+Q_2\\ &=960+2880\\ &\boxed{Q=3840 \text{ Btu}} \end{align*}$

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