Chapter 12, Problem 50A

To determine

Time needed to bring the water at boiling point.

Answer to Problem 50A

Time needed to bring the water at boiling point is $t=367\text{s}$ .

Explanation of Solution

Given:

The given figure is shown below.

  

Power of heater, $P=3\times {10}^2\text{W}$

Mass of cup, $m_1=100\text{g}=0.3\text{kg}$

Mass of water, $m_2=250\text{g}=0.25\text{kg}$

Temperature of water and cup, $T_1=15°\text{C}$

Boiling point of water, $T_2=100°\text{C}$

Formula Used:

Specific heat formula,

  $Q=m_1{C}_1\Delta T_1+m_2{C}_2\Delta T_2$

Here, $Q$ is heat energy, $m_1$ is mass of car, $C_1$ is specific heat of air, and $\Delta T_1$ is change in temperature, $C_2$ is specific heat of water, and $\Delta T_2$ is change in temperature.

  $P=\frac{W}{t}=\frac{Q}{t}$

Here, $Q$ is heat energy and $t$ is time.

Calculation:

As specific heat of water is $C_2=4180\text{J/kg}\cdot \text{K}$ and specific heat of glass is $C_1=840\text{J/kg}\cdot \text{K}$

Substitute the given values to find the value of Q

  $\begin{array}{c}Q=0.3\times 840\times \left(100-15\right)+0.25\times 4180\times \left(100-15\right)\\ =1.1\times {10}^5\text{J}\end{array}$

Now, find the time using power and Q

  $t=\frac{Q}{P}=\frac{1.1\times {10}^5}{300}=367\text{s}$

Conclusion:

Therefore, time needed to bring the water at boiling point is $t=367\text{s}$ .