An engine is working . It takes 100 calories of heat from source and leaves 80 calories of heat to sink . If the temperature of source is `127^(@)` C, then temperature of sink is

Options

A. `147^(@)C`
B. `47^(@)C`
C. `100^(@)C`
D. `47K`

Explanation

To solve the problem step by step, we will use the principles of thermodynamics related to heat engines.

### Step 1: Understand the Given Data
We are given:
- Heat taken from the source (Q1) = 100 calories
- Heat rejected to the sink (Q2) = 80 calories
- Temperature of the source (T1) = 127°C

### Step 2: Calculate the Efficiency of the Engine
The efficiency (η) of a heat engine is given by the formula:
\[ η = 1 - \frac{Q2}{Q1} \]
Substituting the values:
\[ η = 1 - \frac{80}{100} = 1 - 0.8 = 0.2 \]

### Step 3: Convert the Temperature of the Source to Kelvin
To use the efficiency formula in terms of temperatures, we need to convert the temperature from Celsius to Kelvin:
\[ T1 = 127°C + 273 = 400 \text{ K} \]

### Step 4: Use the Efficiency Formula in Terms of Temperature
The efficiency can also be expressed in terms of the temperatures of the source and sink:
\[ η = 1 - \frac{T2}{T1} \]
Substituting the known values:
\[ 0.2 = 1 - \frac{T2}{400} \]

### Step 5: Rearranging the Equation to Solve for T2
Rearranging the equation to find T2:
\[ 0.2 = 1 - \frac{T2}{400} \]
\[ \frac{T2}{400} = 1 - 0.2 = 0.8 \]
\[ T2 = 0.8 \times 400 = 320 \text{ K} \]

### Step 6: Convert T2 Back to Celsius
Now, we convert the temperature of the sink back to Celsius:
\[ T2 = 320 \text{ K} - 273 = 47°C \]

### Final Answer
The temperature of the sink (T2) is **47°C**.

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