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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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