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Chapter 8: Problem 1198

The temperature of sink of Carnot engine is \(27^{\circ} \mathrm{C}\). Efficiency of engine is \(25 \%\) Then find the temperature of source. (A) \(227^{\circ} \mathrm{C}\) (B) \(327^{\circ} \mathrm{C}\) (C) \(27^{\circ} \mathrm{C}\) (D) \(127^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature of the source is approximately \(127^{\circ} \mathrm{C} \) (Option D).

Step by step solution

01

Convert Celsius to Kelvin scale

In order to solve this problem, we need to convert the Celsius temperature to the Kelvin scale. The formula to convert is as follows: K = °C + 273.15 Given, the temperature of the sink of Carnot engine is 27°C. We will convert it to Kelvin: T2 = 27 + 273.15 = 300.15 K
02

Apply the formula for efficiency of Carnot engine

The formula for efficiency of a Carnot engine is Efficiency = ( 1 - (T2/T1) ) x 100%. We are given the efficiency as 25%. We will use this information to find T1: 0.25 = 1 - (T2/T1)
03

Solve for T1

Rearrange the equation to find T1: 0.75 = T2/T1 Now, we'll plug in the value of T2 and solve for T1: 0.75 = 300.15/T1 T1 = 300.15/0.75 T1 = 400.2 K
04

Convert back to Celsius

We've found the temperature of the source in Kelvin. Now, let's convert it back to Celsius using the conversion formula: Temperature in Celsius = Temperature in Kelvin - 273.15 T1_celsius = 400.2 - 273.15 T1_celsius = 127.05°C
05

Check the options

Finally, let's check which option is close to the result we found: (A) 227°C (B) 327°C (C) 27°C (D) 127°C The closest answer is option (D) 127°C.

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Most popular questions from this chapter

An ideal gas heat engine is operating between \(227^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\). It absorbs \(10^{4} \mathrm{~J}\) Of heat at the higher temperature. The amount of heat Converted into. work is \(\ldots \ldots\) J. (A)2000 (B) 4000 (C) 5600 (D) 8000

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