# Applied Sciences

Unit VIII Problem Solving Worksheet

This assignment will allow you to demonstrate the following objectives:

Describe thermodynamic concepts and their applications.

Extend the first law of thermodynamics to various daily life activities.

Identify the maximum efficiency of a heat engine.

Explain the role of latent heat while phases are changing.

Instructions: Choose 8 of the 10 problems below. Show your work in detail. Answer the questions directly in this template. Before doing this, it is highly recommending that you thoroughly review the four examples in the Unit Lesson.

The efficiency of a Carnot engine is e=1-Tc/TH, where Tc is a temperature of the cold reservoir and TH is a temperature of the hot reservoir. What is the condition to have 100% efficiency? Hint: What is the mathematical condition for Tc/TH to be zero.

Suppose the work done to compress a gas is 100 J. If 70 J of heat is lost in the process, what is the change in the internal energy of the gas? Hint: Use the first law of thermodynamics. The internal energy of a system changes due to heat (Q) and work (W): U=Q-W. The change in internal energy is equivalent to the difference between the heat added to the system and the work done by the system. Think if the work done is to the system or by the system. This determines the sign of W.

An engine’s fuel is heated to 2,000 K and the surrounding air is 300 K. Calculate the ideal efficiency of the engine. Hint: The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH : e=W/QH. W=QH – QC, where Qc is the output heat. That is, e=1-Qc/QH =1-Tc/TH, where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir. The unit of temperature must be in Kelvin.

Mr. White claims that he invented a heat engine with a maximum efficiency of 90%. He measured the temperature of the hot reservoir as 100o C and that of cold reservoir as 10o C. Find the error that he made and calculate the correct efficiency. Hint: The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH : e=W/QH. W=QH – QC, where Qc is the output heat. That is, e=1-Qc/QH =1-Tc/TH, where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir. The unit of temperature must be in Kelvin.

How much energy is needed to change 100 g of 0o C ice to 0o C water? The latent heat of fusion for water L=335,000 J/kg. Hint: The heat (Q) used to change from one phase to another phase of the matter is Q=mL, where L is the latent heat. Its unit is J/kg.

It was determined in the 19th century that the normal human body temperature is 98.6o F. A more recent study found that it is 98.2o F. Express the difference in the temperature in Celsius. Hint: Use the converting formula between Fahrenheit and Celsius scales: F=9/5C +32. Be careful about the unit.

Suppose 0.5 kg of blood flows from the interior to the surface of John’s body while he is exercising. The released energy is 2,000 J. The specific heat capacity of blood is 4,186 J/kgo C. What is the temperature difference between when the blood arrives at the body surface and returns back to the interior of the body? Hint: Use the formula regarding heat Q, specific heat capacity c, mass m, and temperature change dT. Q= cm dT. Please look at p.290 in our textbook. Also, review Example 1 with its solution in Study Guide.

A student does 1,000 J of work when she moves to her dormitory. Her internal energy is decreased by 3,000 J. Determine the heat during this process. Does she gain or lose her heat? Hint: Use the first law of thermodynamics. The internal energy of a system changes due to heat Q and work done W: U=Q-W. Also, look at a similar case, Example 3 with its solution in Study Guide.

In a construction site, 2 kg of aluminum shows the increment of temperature by 5oC. Ignoring the work, what is the change in the internal energy of the material? The specific heat capacity of aluminum is 900 J/kg oC. Hint: The internal energy of a system changes due to heat Q and work done W: U=Q-W. If we ignore, the work, the internal energy U is identical to the heat Q of the system. We know that relation between heat Q, specific heat capacity c, mass m, and temperature change dT; Q= cm dT. That is, U=Q=cm dT.

The input heat of a Carnot engine is 3,000 J. The temperature of a hot reservoir is 600 K and that of a cold reservoir is 300 K. What is the work done? Hint: The efficiency e of a Carnot engine is defined as the ratio of the work done, W, by the engine to the input heat QH : e=W/QH. W=QH – QC, where Qc is the output heat. That is, e=1-Qc/QH =1-Tc/TH, where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir. The unit of temperature must be in Kelvin.

Use the formula, e=1-Tc/TH. Please review the Example 4 with its solution in Study Guide. Once you evaluate, you can find the work done of the system using the formula, e=W/QH