A heavy copper pot of mass 2000g ( including the cooper lid) is at temperature of 150°C . 100g of water at 25°C is poured into the pot and the lid of the pot is quickly closed so that no steam can escape. Find the final temperature of the pot and its content. [ specific heat capacity of pot is 0.104 calories/ gram Celsius]
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As, heat transfer takes place from the copper pot(with lid) to the water until same temperature is reached.
let final temperature be T.
T1(copper pot)=150 degrees C
m1=2 kg
c1=0.104 cal g-1 degrees C-1 or 435.344 J kg-1 degrees C-1
Q=mc(T1-T)
T2(water)=25 degrees C
m2=0.1 kg
c2=4200 J kg-1 K-1
m1c1(T1-T)=m2c2(T-T2)
2 X 435.344(150-T)=0.1 X 4200 X (T-25)
870.688(150-T)=420 X (T-25)
T-25=870.688/420(150-T)
T-25=2.073(150-T)
T-25=310.96 - 2.073T
3.073T=285.96
T=93.0556459486
T=93.05 degrees celsius
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