Hot water of mass 0.35kg at 80degrees celsius is poured into a copper cup which is initially at 25 degrees celsius. The mass of the cup is 0.80kg. Assume that no energy is lost to the surroundings.

Give - Specific Heat Capacity of water = 4200Jkg^-1C^-1

Specific Heat Capacity of Copper = 400Jkg^-1C^-1

Question) Find the initial Temperature of water.

Thanks,

Vishnu

Dear Vishnu,

since no energy is lost to surroundings

heat lost by water = heat gained by copper cup

let final common temperature of the mixture = T

therefore,

0.35X4200X(80-T) = 0.8X400X(T-25)

117600-1470T = 320T-8000

=> 117600+8000 = 1470T+320T

125600=1790T

=>T = 125600/1790 = 70.16 degree C

final common temperature of mixture = 70.16 degree C

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