An ideal gas heat engine operates in a Carnot cycle between `227^(@)C and 127^(@)C`. It absorbs `6K cal.` of heat at higher temperature. The amount of heat in `k cal` ejected to sink is

Options

A. 1.6
B. 1.2
C. 4.8
D. 3.5

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
The temperatures given are:
- T1 = 227°C
- T2 = 127°C

To convert these temperatures to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Calculating T1:
\[ T1 = 227 + 273 = 500 \, K \]

Calculating T2:
\[ T2 = 127 + 273 = 400 \, K \]

### Step 2: Use the Carnot engine efficiency relationship
In a Carnot cycle, the relationship between the heat absorbed (Q1) and the heat rejected (Q2) is given by:
\[ \frac{Q2}{Q1} = \frac{T2}{T1} \]

We know:
- Q1 = 6 kcal (heat absorbed at the higher temperature)
- T1 = 500 K
- T2 = 400 K

### Step 3: Rearrange the formula to find Q2
From the relationship, we can express Q2 as:
\[ Q2 = Q1 \cdot \frac{T2}{T1} \]

### Step 4: Substitute the values into the equation
Now substituting the known values into the equation:
\[ Q2 = 6 \, \text{kcal} \cdot \frac{400 \, K}{500 \, K} \]

### Step 5: Simplify the equation
Calculating the fraction:
\[ Q2 = 6 \cdot \frac{400}{500} = 6 \cdot \frac{4}{5} \]

### Step 6: Calculate the final value
Now, calculating the final value:
\[ Q2 = 6 \cdot 0.8 = 4.8 \, \text{kcal} \]

Thus, the amount of heat ejected to the sink is:
\[ \boxed{4.8 \, \text{kcal}} \]

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