Chapter 14, Problem 59GP

To determine

The quantity of water to maintain the desired temperature.

Answer to Problem 59GP

The quantity of water to maintain the desired temperature is $1.1\times {10}^{-6}\text{kg}$ .

Explanation of Solution

Given:

The desired temperature is, $T=35°\text{C}=\left(35°\text{C}+273\right)\text{K}=308\text{K}$ .

Formula Used:

The expression to calculate the quantity of water to maintain the desired temperature is,

  $\dot{m}=\epsilon A\left(\frac{\left(1000\text{W}/{\text{m}}^2\right)\cos \theta -2\sigma \left(T_d^2-T^2\right)}{L_e}\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}\dot{m}=\left[\left(0.85\right)\left(4\times {10}^{-3}{\text{m}}^2\right)\left(\frac{\begin{array}{l}\left(1000\text{W}/{\text{m}}^2\right)\left(\cos 90°\right)-\\ \left(2\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot \text{K}\right)\left({\left(308\text{K}\right)}^4-{\left(293\text{K}\right)}^4\right)\right)\end{array}}{2.45\times {10}^6\text{J}/\text{kg}}\right)\right]\\ =1.1\times {10}^{-6}\text{kg}\end{array}$

Conclusion:

Thus, the quantity of water to maintain the desired temperature is $1.1\times {10}^{-6}\text{kg}$ .