Chapter 6, Problem 6.77E

Using the specific heat data of Table 6.8, calculate the amount of heat (in calories) needed to increase the temperature of the following:

a. $210\text{g}$ of copper from $40.°\text{C}$ to $95.°\text{C}$

b. $150\text{g}$ of mercury from $120.°\text{C}$ to $300.°\text{C}$

c. $2.50\times {10}^3\text{g}$ of helium gas from $250.°\text{C}$ to $900.°\text{C}$

Interpretation Introduction

(a)

Interpretation:

To calculate the heat (in calories) required to increase the temperature of $210\text{g}$ copper from $40.°\text{C}$ to $95.°\text{C}$.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The energy required to change temperature of matter is known as specific heat of matter. The energy required to change a state of matter is known as heat of fusion or vaporization.

Answer to Problem 6.77E

The amount of heat (in calories) required to increase the temperature of $210\text{g}$ copper from $40.°\text{C}$ to $95.°\text{C}$ is $1074.15\text{cal}$.

Explanation of Solution

The formula to calculate amount of heat (in calories) required to increase the temperature is given below as,

$\text{Heat}=\left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)\\ & =& \left(210\text{g}\right)\left(\frac{\text{0}\text{.093}\text{cal}}{\text{g°C}}\right)\left(95°\text{C}-40°\text{C}\right)\\ & =& \left(210\text{g}\right)\left(\frac{\text{0}\text{.093}\text{cal}}{\text{g°C}}\right)\left(55°\text{C}\right)\\ & =& 1074.15\text{cal}\end{array}$

Conclusion

The amount of heat (in calories) required to increase the temperature of $210\text{g}$ copper from $40.°\text{C}$ to $95.°\text{C}$ is $1074.15\text{cal}$.

Interpretation Introduction

(b)

Interpretation:

The amount of heat (in calories) required to increase the temperature of $150\text{g}$ mercury from $120.°\text{C}$ to $300.°\text{C}$ is to be calculated.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The energy required to change temperature of matter is known as specific heat of matter. The energy required to change a state of matter is known as heat of fusion or vaporization.

Answer to Problem 6.77E

The amount of heat (in calories) required to increase the temperature of $150\text{g}$ mercury from $120.°\text{C}$ to $300.°\text{C}$ is $891\text{cal}$.

Explanation of Solution

The formula to calculate amount of heat (in calories) required to increase the temperature is given below as,

$\text{Heat}=\left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)\\ & =& \left(150\text{g}\right)\left(\frac{\text{0}\text{.033}\text{cal}}{\text{g°C}}\right)\left(300°\text{C}-120°\text{C}\right)\\ & =& \left(150\text{g}\right)\left(\frac{\text{0}\text{.033}\text{cal}}{\text{g°C}}\right)\left(180°\text{C}\right)\\ & =& 891\text{cal}\end{array}$

Conclusion

The amount of heat (in calories) required to increase the temperature of $150\text{g}$ mercury from $120.°\text{C}$ to $300.°\text{C}$ is $891\text{cal}$.

Interpretation Introduction

(c)

Interpretation:

The amount of heat (in calories) required to increase the temperature of $2.50\times {10}^3\text{g}$ helium gas from $250.°\text{C}$ to $900.°\text{C}$ is to be calculated.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The energy required to change temperature of matter is known as specific heat of matter. The energy required to change a state of matter is known as heat of fusion or vaporization.

Answer to Problem 6.77E

The amount of heat (in calories) required to increase the temperature of $2.50\times {10}^3\text{g}$ helium gas from $250.°\text{C}$ to $900.°\text{C}$ is $2.03\times {10}^6\text{cal}$.

Explanation of Solution

The formula to calculate amount of heat (in calories) required to increase the temperature is given below as,

$\text{Heat}=\left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)\\ & =& \left(2.50\times {10}^3\text{g}\right)\left(\frac{\text{1}\text{.25}\text{cal}}{\text{g°C}}\right)\left(900°\text{C}-250°\text{C}\right)\\ & =& \left(2.50\times {10}^3\text{g}\right)\left(\frac{\text{1}\text{.25}\text{cal}}{\text{g°C}}\right)\left(650°\text{C}\right)\\ & =& 2.03\times {10}^6\text{cal}\end{array}$

Conclusion

The amount of heat (in calories) required to increase the temperature of $2.50\times {10}^3\text{g}$ helium gas from $250.°\text{C}$ to $900.°\text{C}$ is $2.03\times {10}^6\text{cal}$.