Physics, asked by husainaltaph3, 3 months ago

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]​

Answers

Answered by Anonymous
3

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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