Question

1. an ideal carnot heat engine has an efficiency of 0.600. if it operates between a deep lake with a constant temperature of 21°c and a hot reservoir, what is the temperature of the hot reservoir? (t_{kelvin}=t_{c}+273.15)

-a. 735 k
b. 490 k
c. 470 k
d. 784 k

1. a heat engine with an efficiency of 30% performs 2500 j of work. how much heat, approximately, is released to the lower temperature reservoir?

-a. 5800 j
b. 8300 j
c. 750 j
d. 1400 j

1. a stretched string shows four equal segments in a standing wave at a frequency of 480 hz. what frequency should be used to create a standing wave with five equal segments?

a. 600 hz
b. 360 hz
c. 240 hz
d. 120 hz

Explanation:

Question 7

Step1: Convert cold - reservoir temperature to Kelvin

The formula for converting Celsius to Kelvin is \(T = T_{C}+273.15\). Given \(T_{C}=21^{\circ}C\), then \(T_{C}=21 + 273.15=294.15\ K\)

Step2: Use the Carnot efficiency formula

The Carnot efficiency formula is \(\eta=1-\frac{T_{C}}{T_{H}}\), where \(\eta = 0.600\). Rearranging for \(T_{H}\) gives \(T_{H}=\frac{T_{C}}{1 - \eta}\)
Substitute \(T_{C}=294.15\ K\) and \(\eta = 0.600\) into the formula: \(T_{H}=\frac{294.15}{1 - 0.600}=\frac{294.15}{0.400}=735.375\approx735\ K\)

Step1: Use the efficiency formula to find heat input

The efficiency formula is \(\eta=\frac{W}{Q_{H}}\), where \(\eta = 0.30\) and \(W = 2500\ J\). Rearranging for \(Q_{H}\) gives \(Q_{H}=\frac{W}{\eta}=\frac{2500}{0.30}\approx8333.33\ J\)

Step2: Use the first - law of thermodynamics for heat engines

The first - law of thermodynamics for heat engines is \(Q_{H}=W + Q_{C}\). Rearranging for \(Q_{C}\) gives \(Q_{C}=Q_{H}-W\)
Substitute \(Q_{H}\approx8333.33\ J\) and \(W = 2500\ J\) into the formula: \(Q_{C}=8333.33-2500 = 5833.33\approx5800\ J\)

Step1: Recall the formula for the frequency of standing waves on a string

The frequency of standing waves on a string is \(f_{n}=\frac{n}{2L}\sqrt{\frac{T}{\mu}}\), where \(n\) is the number of segments, \(L\) is the length of the string, \(T\) is the tension, and \(\mu\) is the linear mass density. For a given string (\(L\), \(T\), \(\mu\) constant), \(f\propto n\)

Step2: Set up a proportion

We know that \(f_{4}=480\ Hz\) (when \(n = 4\)). Let \(f_{5}\) be the frequency when \(n = 5\). Using the proportion \(\frac{f_{4}}{f_{5}}=\frac{4}{5}\), then \(f_{5}=\frac{5}{4}f_{4}\)
Substitute \(f_{4}=480\ Hz\) into the formula: \(f_{5}=\frac{5}{4}\times480 = 600\ Hz\)

Answer:

A. 735 K

Question 8