33.The heat capacity of (the interior of) a refrigerator is 4.2kJ/K.The minimum work that must be done to lower the internal temperature from 18^(@)C to 17^(@)C when the outside temperature is 27^(@)C will be (a) 2.20kJ (b) 0.80kJ (c) 0.30kJ (d) 0.14kJ?

Question

Answer

Given:
Heat capacity of the interior of a refrigerator (C) = 4.2 kJ/K
Initial temperature (T1) = 18°C
Final temperature (T2) = 17°C
Outside temperature (Tout) = 27°C

To find:
The minimum work required to lower the internal temperature of the refrigerator from 18°C to 17°C.

Solution:
The minimum work required to lower the temperature can be calculated using the formula:

W = C * ΔT

where W is the work done, C is the heat capacity, and ΔT is the change in temperature.

Step 1: Convert the temperatures from Celsius to Kelvin.
The temperature in Kelvin is given by the formula:

T(K) = T(°C) + 273

Initial temperature in Kelvin (T1) = 18°C + 273 = 291 K
Final temperature in Kelvin (T2) = 17°C + 273 = 290 K
Outside temperature in Kelvin (Tout) = 27°C + 273 = 300 K

Step 2: Calculate the change in temperature.
ΔT = T2 - T1 = 290 K - 291 K = -1 K

Step 3: Calculate the minimum work required.
W = C * ΔT = 4.2 kJ/K * (-1 K) = -4.2 kJ

Step 4: Convert the work to a positive value.
Since work cannot be negative, we take the absolute value of the calculated work.

W = |-4.2 kJ| = 4.2 kJ

Step 5: Compare the calculated work with the given options.
The minimum work required is 4.2 kJ.

Comparing with the given options:
(a) 2.20 kJ
(b) 0.80 kJ
(c) 0.30 kJ
(d) 0.14 kJ

Answer:
The minimum work that must be done to lower the internal temperature from 18°C to 17°C when the outside temperature is 27°C is 4.2 kJ, which is not equal to any of the given options.

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