" The temperature inside and outside of refrigerator are 260 K and 315 K respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work done"
Added by Julia B.
Step 1
Given that Q1 = 473 J (as calculated in the Explanation) and W = 100 J (given in the question), we can calculate Q2: Q2 = 473 J - 100 J Q2 = 373 J Show more…
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Dharmendra M.
The inside and outside temperatures of a refrigerator are $273 \mathrm{~K}$ and $303 \mathrm{~K}$ respectively. Assuming that refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be: (a) $10 \mathrm{~J}$ (b) $20 \mathrm{~J}$ (c) $30 \mathrm{~J}$ (d) $50 \mathrm{~J}$
The temperatures of inside and outside of a refrigerator are $273 \mathrm{~K}$ and $303 \mathrm{~K}$ respectively. Assuming that the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surroundings will be nearly (a) $10 \mathrm{~J}$ (b) $20 \mathrm{~J}$ (c) $30 \mathrm{~J}$ (d) $50 \mathrm{~J}$
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