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A piece of copper ball of mass 20g at 200°is place in a copper calorimeter of 60g containing 50g of water at 39°.ignoring heat calculate the final steady temperature of mixture (specific heat capacity water=4400J/Kg/°k
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Answered by
Anonymous
You did not say what the heat capacity of copper is.
I am not about to look it up.
Call it Kc Joules / GRAM deg C (convert to grams if your table is for Kg)
You have 20 g of Cu at 200
You have 60 g of Cu at 39
You have 50 g of H2O at 39 Kw = 4.400 J/ GRAM deg K or C
final temp = T
heat out = Kc (20) (200-T)
heat in = 4.4 * 50 (T-39) + Kc *60 *(T-39)
heat out = heat in, solve for T
I am not about to look it up.
Call it Kc Joules / GRAM deg C (convert to grams if your table is for Kg)
You have 20 g of Cu at 200
You have 60 g of Cu at 39
You have 50 g of H2O at 39 Kw = 4.400 J/ GRAM deg K or C
final temp = T
heat out = Kc (20) (200-T)
heat in = 4.4 * 50 (T-39) + Kc *60 *(T-39)
heat out = heat in, solve for T
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