   # Suppose the same process of adding energy to the ice cube is performed as discussed above with regard to Figure 19.3, but instead we graph the internal energy of the system as a function of energy input. What would this graph look like? ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278

#### Solutions

Chapter
Section ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553278
Chapter 19.3, Problem 19.2QQ
Textbook Problem
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## Suppose the same process of adding energy to the ice cube is performed as discussed above with regard to Figure 19.3, but instead we graph the internal energy of the system as a function of energy input. What would this graph look like?

To determine

To draw: The graph for the internal energy of the system as a function of energy input.

### Explanation of Solution

Given info: The mass of ice is 1.00g Initial temperature of the ice is 30.0°C and the temperature is raised to 120°C . The temperature variation with respect to the energy added in the system is as shown below.

The given figure is shown below.

Figure (1)

Calculate energy for each part of the graph.

For part A

Q=mcΔT (1)

Here,

Q is the amount of energy added.

m is the mass of the sample.

ΔT is the change of temperature.

c is the specific heat capacity of ice.

Substitute 2090J/kg°C for c and 1.00g for m and 30.0°C for ΔT in equation (1).

Q=mcΔT=1.00g×2090J/kg-°C×30.0°C=1.001kg1000g×2090J/kg-°C×30.0°C=62.7 J

Therefore the total internal energy of the system is 62.7J for part A of the curve.

For part B of the given curve there is change of phase from ice to water so

The formula for Energy added is

Q=Lfm (2)

Here,

Lf is the latent heat

Substitute 3.33×105J/kg for Lf and 1.00g for m .

Q=3.33×105J/kg×1.001kg1000g=333J

Thus The total internal energy of the system 333J+62.7J=396J

For part C the energy added is

Q=mcΔT

Substitute 4.19×103J/kg°C for c and 1.00g for m and 100.0°C for ΔT

Q=1.00g×1kg1000g×4

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