Question

# Heat transfer in calories is given by$Q = m c \Delta T,$where m is mass in grams, c is specific heat capacity in$\text { cal/g } ^ { \circ } \mathrm { C } , \text { and } \Delta T \text { is in } ^ { \circ } \mathrm { C .}$What mass of water will give up 240 calories when its temperature drops from$80^\circ C \text {to} 68^\circ C?$

Solution

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The following are given in the problem;

\begin{align*}Q&=240\ \text{cal}\\T_{1}&=80^{\circ}C\\T_{2}&=68^{\circ}\end{align*}

Now the problem requires us to determine the mass it would take to heat up water, This can be solved by applying the general formula for the amount of heat energy which should be;

\begin{align*}\dot Q&=mc_{p}\Delta T\end{align*}

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