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Question
A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains and equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J kg−1 °C−1.
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Solution
Given:-
Mass of iron = 100 g
Water equivalent of calorimeter = 10 g
Mass of water = 240 gm
Let the temperature of surface be θ°C.
Specific heat capacity of iron = 470 J kg−1 °C−1
Total heat gained = Total heat lost
`rArr100/1000xx470xx(theta-60^o)=(240+10)/1000xx4200xx(60-20)`
`rArr47theta-47xx60=25xx42xx40`
`rArrtheta=(42000+2820)/47=44820/47`
= 953.61°C
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