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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Physics (PH)
40 kcal of heat transferred from a hot body (100°c) to surroundings (27 c). change in entropy of universe is cal k1
40 kcal of heat transferred from a hot body (100°C) to surroundings (27 C). Change in entropy of universe is cal k1
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Ruby negi
The hot reservoir has a loss of entropy ΔS1=−Q1/T1 because heat transfer occurs out of it (remember that when heat transfers out, then Q has a negative sign). The surrounding has a gain of entropy ΔS2=Q1/T2 because heat transfer occurs into it. (We assume the reservoirs are sufficiently large that their temperatures are constant.) So the total change in entropy of universe is ΔS= ΔS1 + ΔS2..
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Ruby negi
Q1 is the heat that lost by hot body and Q1 is the heat that is gain by surrounding. that's why we take negative of Q1 in case of hot body (as it loss heat)and positive of Q1 in case of surrounding... u can convert it into joule..
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Dhairya sharma
for a system which is loosing heat Q1 will be negative and for surrounding which is taking it....i.e.Q2 will be positive
hope u will get it
yes, mam
is there any doubt