Chapter 13, Problem 65P

To determine

To Calculate: The temperature of the water at which water start boil .

Answer to Problem 65P

The temperature of the water at which water start boil is ${91}^0\text{C}$ .

Explanation of Solution

Given:

Pressure = $0.72\text{atm}$

Calculation:

At boiling temperature, the saturated pressure and vapor pressure both are equal.At $0.72\text{atm}$

Temperature is lie between the $100{}^0\text{C}$ and $90{}^0\text{C}$ .

The temperature at $0.72\text{atm}$ cannot be calculated for the this value. So it can be calculated by the linear interpolation

  $\frac{{\left(100-90\right)}^0\text{C}}{\left(10.1-7.01\right)\times {10}^4\text{Pa}}={(3.236\times {10}^{-4})}^0\text{C/Pa}$

Temperature corresponding to $7.27\times {10}^4\text{Pa}$ is

  $\begin{array}{l}{90}^0\text{C}+[(7.27-7.01)\times {10}^4\text{Pa}]\left(3.236\times {10}^{-4}{}^0\text{C/Pa}\right)\\ {90.8}^0\text{C}\approx {91}^0\text{C}\end{array}$

Conclusion:

The temperature at $0.72\text{atm}$ is ${91}^0\text{C}$ .