The morning before a party you take a 95 kg watermelon out o

The morning before a party, you take a 9.5 kg watermelon out of the refrigerator. When you set it on the counter, the watermelon is initially at 4 degree C and the kitchen is at room temperature (20 degree C). After a while, the watermelon has completely warmed up. What is the total entropy change associated with this heat transfer process? kJ/K

Solution

Given mass of water melon 9.5kg

Specific heat of water melon c= 3.96 kj / kg k

T1 = 4°c = 277K

T2 = 20°c = 293K

Total entropy change S = mxc ln ( T2/T1)

S = 9.5 x3.96 ln (293/277)

S = 2.1125 kj /k