Chapter 21, Problem 61A

What will be the final temperature of 100 g of $20°C$ water when 100 g of $40°C$ iron nails are submerged in it? (The specific heat of iron is $0.12cal/g°C$.)

To determine

Final temperature of mixture.

Answer to Problem 61A

Final temperature is, $T=22.14°\text{C}$ .

Explanation of Solution

Given:

Mass of water at $20°\text{C}$ is, $m_1=100\text{ g }$

Mass of iron nails at $40°\text{C}$ is, $m_2=100\text{ g }$

Specific heat of iron nails, $c_2=0.12\text{ cal/g}°\text{C}$

Formula used:

  $Q=mc\Delta t$

Where, Q is amount of heat generated, m is mass, c is specific heat and $\Delta t$ is change in temperature.

Calculation:

According to heat transfer heat lost by hot water is equal to heat gained by cold water.

  $m_1{c}_1\Delta t_1=m_2{c}_2\Delta t_2$

Specific heat of water is, $c_1=1\text{ cal/g}°\text{C}$

Let, final temperature is $T°\text{C}$ .

Therefore,

  $\begin{array}{c}100\times 1\times \left(T-20\right)=100\times 0.12\times \left(40-T\right)\\ 100T-2000=480-12T\\ 112T=2480\\ T=\frac{2480}{112}\\ =22.14°\text{C}\end{array}$

Conclusion:

Therefore, final temperature of mixture is $22.14°\text{C}$ .