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?

Q: Expert Answer

To determine To draw: The graph for the internal energy of the system as a function of energy input. Explanation Given info: The mass of ice is 1.00 g 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 = m cΔT (1) Here, Q is the amount of energy added. m is the mass of the sample. Δ T is the change of temperature. c is the specific heat capacity of ice. Substitute 2090 J / kg ⋅ °C for c and 1.00 g for m and 30 .0°C for Δ T in equation (1). Q = m c Δ T =1 .00 g×2090 J / kg-°C × 30 .0 °C =1 .00 g× 1 kg 1000 g ×2090 J / kg-°C × 30 .0 °C = 62 .7 J Therefore the total internal energy of the system is 62 .7 J 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 = L f m (2) Here, L f is the latent heat Substitute 3.33 × 10 5 J / kg for L f and 1.00 g for m . Q =3 .33×10 5 J / kg ×1 .00 g× 1 kg 1000 g =333 J Thus The total internal energy of the system 333 J+62 .7 J=396 J For part C the energy added is Q = m cΔT Substitute 4.19 × 10 3 J / kg ⋅ °C for c and 1.00 g for m and 100 .0°C for Δ T Q = 1.00 g × 1 kg 1000 g × 4

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Answer 1

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Physics for Scientists and Enginee...

10th Edition

Raymond A. Serway + 1 other

Publisher: Cengage Learning

ISBN: 9781337553278

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Chapter

Section

Problem 19.2QQ

Answer 2

BuyFind

Physics for Scientists and Enginee...

10th Edition

Raymond A. Serway + 1 other

Publisher: Cengage Learning

ISBN: 9781337553278

Solutions

Chapter

Section

Problem 19.2QQ

Answer 3

Solutions

Chapter

Section

Problem 19.2QQ

Answer 4

Chapter

Section

Answer 5

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 1.00 g 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 1.00 g for m and 30.0°C for ΔT in equation (1).

Q=mcΔT=1.00 g×2090 J/kg-°C×30.0 °C=1.00 g×1 kg1000 g×2090J/kg-°C×30.0 °C=62.7 J

Therefore the total internal energy of the system is 62.7 J 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 1.00 g for m .

Q=3.33×105J/kg×1.00 g×1 kg1000 g=333 J

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

For part C the energy added is

Q=mcΔT

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

Q=1.00 g×1 kg1000 g×4

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...

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Physics for Scientists and Enginee...