Chapter 13, Problem 28P

(a)

To determine

To Convert: The $°\text{C}$ into the Kelvin scale.

Answer to Problem 28P

  $\text{43}00\text{K}$ and $15\times {10}^6\text{K}$

Explanation of Solution

Given:

The temperature of the Earth = $4000°\text{C}$ .

The temperature of the sun = $15\times {10}^6°\text{C}$ .

Formula used:

Using the relation

  $T(\text{K})=T(°\text{C})+273.15$

Calculation For Earth

  $\begin{array}{c}T(\text{K})=T(°\text{C})+273.15\\ T(\text{K})=T(4000+273.15)\\ =4273.15\text{K}\approx 4300\text{K}\end{array}$

For sun

  $\begin{array}{c}T(\text{K})=T(°\text{C})+273.15\\ =15\times {10}^6+273.15\\ =15\times {10}^6\text{K}\end{array}$

Conclusion:

The temperature of the earth and sun is approximately is $\text{43}00\text{K}$ and $15\times {10}^6\text{K}$ respectively

(b)

To determine

To Calculate: The value of % error of the earth and sun without change the temperature in Kelvin.

Answer to Problem 28P

The value of % error of the earth and sun without change the temperature in Kelvin is

  $\approx 7\%$ and $2\times {10}^{-3}\%$ respectively.

Explanation of Solution

Given:

The temperature of the Earth = $4000°\text{C}$ .

The temperature of the sun = $15\times {10}^6°\text{C}$ .

Calculation:

  $\begin{array}{l}\%\text{Error}=\frac{\Delta T}{T(\text{K})}\times 100\\ =\frac{273}{T(\text{K})}\times 100\\ =\frac{273}{4000}\times 100\\ \approx 7\%\\ \frac{273}{15\times {10}^6}\times 100\\ =1.8\times {10}^{-3}\approx 2\times {10}^{-3}\%\end{array}$

Conclusion:

The % error for the earth is $7\%$ and for the sun is $2\times {10}^{-3}\%.$