Chapter 21, Problem 27A

What mass of water will give up 240 calories when its temperature drops from $80°C$ to $68°C$?

To determine

Mass of water.

Answer to Problem 27A

Mass of water is $m=20\text{g}$

Explanation of Solution

Given:

Initial temperature of water is $80°\text{C}$

Final temperature of water is $68°\text{C}$

The number of calories given up by water is $Q=240\text{cal}$

Formula Used:

The heat transfer in calories is given by,

  $Q=mC\Delta T$

Where, $m$ is mass in grams, $C$ is specific heat capacity in $\text{cal/g}°\text{C}$ and $\Delta T$ is change in temperature in $°\text{C}$ .

Calculation:

The change in temperature is,

  $\Delta T=80°\text{C}-68°\text{C}=12°\text{C}$

As, Specific heat capacity of water is, $1\text{cal/g}°\text{C}$

Now, putting the values in heat transfer equation

  $\begin{array}{l}Q=mC\Delta T\\ 240=m\times \left(1\text{cal/g}°\text{C}\right)\times \left(12°\text{C}\right)\\ \Rightarrow m=\frac{240}{12}\text{g}\\ \Rightarrow m=20\text{g}\end{array}$

Conclusion:

Therefore, the mass of water is $m=20\text{g}$ .