Question

A refrigerator consumes an average 35 W power to operate between temperature -10\(^{\circ}\)C to 25\(^{\circ}\)C. If there is no loss of energy then how much average heat per second does it transfer ?

• A. 35 J/s
• B. 263 J/s
• C. 298 J/s
• D. 350 J/s

Hint:

Always convert temperatures to Kelvin for thermodynamic calculations involving ratios, like in the Carnot efficiency or COP formulas. Remember that for a refrigerator, COP relates the heat removed from the cold side (\(Q_L\)) to the work input (W).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the power consumption and operating temperatures of a refrigerator. Assuming it operates ideally (like a Carnot heat pump in reverse), we need to find the rate at which it extracts heat from the cold reservoir.
Step 2: Key Formula or Approach:
1. The coefficient of performance (COP) of an ideal refrigerator is given by \( \text{COP} = \frac{T_L}{T_H - T_L} \), where \(T_L\) and \(T_H\) are the absolute temperatures of the cold and hot reservoirs, respectively.
2. The COP is also defined as the ratio of the heat extracted from the cold reservoir (\(Q_L\)) to the work done (\(W\)) on the refrigerator: \( \text{COP} = \frac{Q_L}{W} \).
3. Power (P) is the work done per unit time, \( P = \frac{W}{t} \). The rate of heat transfer is \( \frac{Q_L}{t} \).
Step 3: Detailed Explanation:
First, convert the temperatures to Kelvin (the absolute temperature scale):
Low temperature (inside the refrigerator), \( T_L = -10^\circ C + 273 = 263 \) K.
High temperature (outside environment), \( T_H = 25^\circ C + 273 = 298 \) K.
Next, calculate the ideal COP:
\[ \text{COP} = \frac{T_L}{T_H - T_L} = \frac{263}{298 - 263} = \frac{263}{35} \] The power consumed is the rate of work done, \( P = \frac{W}{t} = 35 \) W = 35 J/s.
The rate of heat transfer from the cold reservoir is \( \frac{Q_L}{t} \).
From the definition of COP:
\[ \text{COP} = \frac{Q_L}{W} = \frac{Q_L/t}{W/t} = \frac{\text{Rate of heat transfer}}{\text{Power}} \] We can rearrange this to find the rate of heat transfer:
\[ \frac{Q_L}{t} = \text{COP} \times P \] \[ \frac{Q_L}{t} = \left(\frac{263}{35}\right) \times 35 \text{ J/s} \] \[ \frac{Q_L}{t} = 263 \text{ J/s} \] Step 4: Final Answer:
The refrigerator transfers an average heat of 263 J/s. This corresponds to option (B).