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gogeta2006
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Homework Statement
50ml of water at 49.6 C were mixed with 50ml of water at 25.1 C in a calorimeter also at 25.1 C. The final temperature was 30.1 C Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter.
Homework Equations
Density of water = 1.00 g/mL
Specific heat capacity = 4.18 J / g * K
The Attempt at a Solution
q (heat given up by water) = 50ml * (49.6-30.1)
= 840 cal
q (heat absorbed by cold water) = 50ml (30.1-25.1)
= 250 cal
Heat absorbed by calorimeter = 250 + 840 = 1090 cal
Ccal = qcal / delta T
= 590 / (30.1-25.1)
= 118 K
The answer is supposed to be 493.24 J/K ... but i am not getting that.
Please someone please show me how to correct this.
Thank you.