Chapter 10, Problem 1SP

(a)

To determine

The temperature change of the object due to the addition of given amount of heat, in units of degree Fahrenheit.

Answer to Problem 1SP

The temperature change of the object due to the addition of heat, in units of degree Fahrenheit is $90\text{°F}$.

Explanation of Solution

Given info: The initial temperature of the object is $30°\text{C}$ and its final temperature is $80°\text{C}$.

Write the expression to find the change in temperature of the object.

$\Delta T=T_2-T_1$ (1)

Here,

$T_1$ is the initial temperature

$T_2$ is the final temperature

Write the formula for converting temperature in degree Celsius to degree Fahrenheit.

$T\left(\text{F}\right)=\frac{9}{5}T\left(\text{°C}\right)+32$ (2)

Here,

$T\left(\text{F}\right)$ is the temperature in degree Fahrenheit

$T\left(\text{°C}\right)$ is the temperature in degree Celsius

Use equation (2) in (1).

$\begin{array}{c}\Delta T\left(\text{°F}\right)=\left(\frac{9}{5}{T}_2\left(\text{°C}\right)+32\right)-\left(\frac{9}{5}{T}_1\left(\text{°C}\right)+32\right)\\ =\frac{9}{5}{T}_2\left(\text{°C}\right)-\frac{9}{5}{T}_1\left(\text{°C}\right)\\ =\frac{9}{5}\left(T_2\left(\text{°C}\right)-T_1\left(\text{°C}\right)\right)\end{array}$ (3)

Substitute $80\text{°C}$ for $T_2$ and $30\text{°C}$ for $T_1$ in equation (3) to find the change in temperature of the object in units of degree Fahrenheit.

$\begin{array}{c}\Delta T\left(\text{°F}\right)=\frac{9}{5}\left(80\text{°C}-30\text{°C}\right)\\ =90\text{°F}\end{array}$

Conclusion:

Therefore, the temperature change of the object in degree Fahrenheit is $90\text{°F}$.

(b)

To determine

The temperature change of the object in Kelvin scale.

Answer to Problem 1SP

The temperature change of the object in Kelvin scale is $50\text{K}$.

Explanation of Solution

The change in temperature of the object in Celsius scale is obtained as $50\text{°C}$.

Write the formula for converting the temperature in Celsius scale to Kelvin scale.

$T\left(\text{K}\right)=T\left(\text{°C}\right)+273$ (4)

Here,

$T\left(\text{K}\right)$ is the temperature in Kelvin scale.

Use equation (4) in (1).

$\begin{array}{c}\Delta T\left(\text{K}\right)=\left(T_2\left(\text{°C}\right)+273\right)-\left(T_1\left(\text{°C}\right)+273\right)\\ =T_2\left(\text{°C}\right)-T_1\left(\text{°C}\right)\end{array}$ (5)

Substitute $80\text{°C}$ for $T_2$ and $30\text{°C}$ for $T_1$ to find the change in temperature in the units of Kelvin.

$\begin{array}{c}\Delta T\left(\text{K}\right)=\text{80°C}-\text{30°C}\\ =50\text{K}\end{array}$

Conclusion:

Therefore, The temperature change of the object in Kelvin scale is $50\text{K}$.

(c)

To determine

Whether there is any difference in the numerical value of heat capacity expressed in $\text{cal}/\text{g}\cdot °\text{C}$ from the one expressed in $\text{cal}/\text{g}\cdot \text{K}$.

Answer to Problem 1SP

There is no difference in the numerical value of heat capacity expressed in $\text{cal}/\text{g}\cdot °\text{C}$ from the one expressed in $\text{cal}/\text{g}\cdot \text{K}$.

Explanation of Solution

The specific heat of a substance is the amount of heat energy required to raise the temperature of a substance by $1\text{°C}$ or $1\text{K}$. It is the characteristic of a particular material.

Write the expression for the temperature difference in the units of Kelvin scale.

$\Delta T\left(\text{K}\right)=T_2\left(\text{K}\right)-T_1\left(\text{K}\right)$ (6)

Use $T\left(\text{K}\right)=T\left(\text{°C}\right)+273$ in the right hand side of equation (6).

$\begin{array}{c}\Delta T\left(\text{K}\right)=\left(T_2\left(\text{°C}\right)+273\right)-\left(T_1\left(\text{°C}\right)+273\right)\\ =T_2\left(\text{°C}\right)-T_1\left(\text{°C}\right)\\ =\Delta T\left(\text{°C}\right)\end{array}$

Thus, the magnitude of change in temperature is same irrespective of the measurement in Kelvin scale or Celsius scale. This implies that the increase of $1\text{°C}$ is equivalent to the increase of $1\text{K}$.

Therefore, the heat energy required to raise the temperature of a substance by $1\text{°C}$ will be equal to the raise in temperature by $1\text{K}$. In other words, the magnitude of specific heat will be same when it is written in units of $\text{cal}/\text{g}\cdot °\text{C}$ or $\text{cal}/\text{g}\cdot \text{K}$.

Conclusion:

Therefore, there is no difference in the numerical value of heat capacity expressed in $\text{cal}/\text{g}\cdot \text{C}°$ from the one expressed in $\text{cal}/\text{g}\cdot \text{K}$.