250.0 g of H2O at 82.0 C was added to a calorimeter containing 400.0 g of H2O at 18.0 C. The final temp of the mixture was determined to be 42.4 C. Calculate the heat capacity of the calorimeter.
Answer:
Heat capacity is the amount of heat to raise the mass of a substance by 1 degree Celsius.
Use this equation: C = q x m x deltaT
C = specific heat
q = amount of heat
m = mass of object
deltaT = change in temperature
the specific heat of water is 4.184kJ/mol
4.184kJ = 400g x q x 24.4
Just solve for q which is the amount of heat absorbed. :)
250g x 4.184J/g/°C x 39.6°C ΔT
= 41,421.6 J = heat lost by hot water.
400g x 4.184J/g/°C x 24.4°C ΔT
= 40,835.8 J = heat gained by the cold water.
Difference to the calorimeter
= 41,421.6 - 40,835.8 = 585.76 J
Heat capacity of the calorimeter...
= 585.76 J ÷ mass of calorimeter in grams. = J/g/°C
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