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Question

What Btu\mathrm{Btu} of heat are released when 20.0 lb20.0 \mathrm{~lb} of water at 80°F80{\degree} \mathrm{F} is cooled to 32°F32{\degree} \mathrm{F} and then frozen in an ice plant?

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Answered 2 years ago
Answered 2 years ago
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Given:\textbf{Given:}

w=20 lbw=20 \text{ lb}

T1=32FT_1=32 ^\circ\text{F}

T2=80FT_2=80 ^\circ\text{F}

Lf=144 Btu lbL_f=144 \text{ } \dfrac{ \text{Btu}}{ \text{ lb}}

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

We first calculate the heat required to cool water to 32F32 ^ \circ \text {F} using the heat formula:

Q1=wcΔTQ_1=w\cdot c \cdot \Delta T

By entering the given data we get:

Q1=wc(T2T1)=201 (8032)Q1=960 Btu\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:

Lf=Q2wL_f=\dfrac{Q_2}{w}

We express the heat and include the given data:

Q2=Lfw=14420Q2=2880 Btu\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:

Q=Q1+Q2=960+2880Q=3840 Btu\begin{align*} Q&=Q_1+Q_2\\ &=960+2880\\ &\boxed{Q=3840 \text{ Btu}} \end{align*}

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