Chapter 23, Problem 16A

(a)

To determine

The energy in calories needed to raise the temperature of 1 gram of water by 1⁰C

Answer to Problem 16A

The heat needed is 1 Calorie.

Explanation of Solution

Given info:

The amount of water m = 1 g

The change in temperature of water

  ${\text{ΔT=1}}^{\text{0}}\text{C}$

The specific capacity of water C_w= 1 Cal/g ⁰C

Let Q is the heat energy absorbed by water.

Formula used:

  ${\text{Q = mC}}_{\text{w}}\text{ΔT}$

Calculation:

Substituting the given values, we get

  $\begin{array}{l}\text{Q =}\left(\text{1g}\right)\left({\text{1 Cal/g}}^{\text{0}}\text{C}\right)\left(1^{\text{0}}\text{C}\right)\\ \text{Q = 1 Cal}\end{array}$

Conclusion:

Thus, the energy absorbed by 1g water to raise the temperature of 1 gram of water by 1⁰C is

1 calorie.

(b)

To determine

The energy in calories needed to melt1 gram of water at 0⁰C

Answer to Problem 16A

The heat needed is 80 Calories.

Explanation of Solution

Given info:

The amount of water m = 1 g

The latent heat of fusion of water L_f= 80 Cal/g

Let Q is the heat needed.

Formula used:

  ${\text{Q = mL}}_{\text{f}}$

Calculation:

Substituting the given values, we get

  $\begin{array}{l}\text{Q =}\left(\text{1g}\right)\left(\text{80 Cal/g}\right)\\ \text{Q = 80 Cal}\end{array}$

Conclusion:

Thus, the energy in calories needed to melt 1 gram of ice 0⁰C is 80 calories.

(c)

To determine

The energy in calories needed to vaporize1 gram of water at 100⁰C

Answer to Problem 16A

The heat needed is 540 Calories.

Explanation of Solution

Given info:

The amount of water m = 1 g

The latent heat of vaporization of water L_v= 540 Cal/g

Let Q is the heat needed.

Formula used:

  ${\text{Q = mL}}_{\text{v}}$

Calculation:

Substituting the given values, we get

  $\begin{array}{l}\text{Q =}\left(\text{1g}\right)\left(\text{540 Cal/g}\right)\\ \text{Q = 540 Cal}\end{array}$

Conclusion:

Thus, the energy in calories needed to vaporize 1 gram of water at 100⁰C is 540 calories.