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Step-by-Step Solution
Step 1: Understand the Problem
The question involves a refrigerator operating between two temperatures, 4°C and 30°C. We need to determine the power required to remove 600 calories of heat per second, given that 1 cal = 4.2 J.
Step 2: Convert Temperatures to Kelvin
In thermodynamics, temperatures should be in Kelvin. Thus,
$T_L = 4^\circ C + 273 = 277\,K$
$T_H = 30^\circ C + 273 = 303\,K$
Step 3: Convert Heat Removed to Joules per Second
We are removing 600 calories of heat every second. Since 1 cal = 4.2 J,
$L_Q = 600 \times 4.2 = 2520\,\text{J/s}$
This is the rate of heat removal from the refrigerated space ($Q_L$).
Step 4: Write the Formula for Coefficient of Performance (COP)
For a refrigerator, the Coefficient of Performance (COP) is given by:
$COP = \frac{T_L}{T_H - T_L}$
Also, by definition:
$COP = \frac{Q_L}{Power} \quad \Longrightarrow \quad Power = \frac{Q_L}{COP}$
Step 5: Substitute and Calculate the Power Required
Using $COP = \frac{T_L}{T_H - T_L}$, we get:
$COP = \frac{277}{303 - 277} = \frac{277}{26}
$
Therefore, the power required is:
$Power = \frac{Q_L}{COP} = \frac{2520}{\frac{277}{26}} = 2520 \times \frac{26}{277}
$
Calculating this:
$Power \approx 236.5\,\text{W}
$
Step 6: Final Answer
The power required by the refrigerator is approximately 236.5 W.
