Question

Calculate the specific heat of a metal from the following experimental data. 75.0 ml cold water is taken in a calorimeter. The initial temp of the water in the calorimeter is 21.2 degrees C. To the calorimeter containing cold water 29.458 g metal at 98.9 degrees C is added. The final temperature of the contents of the calorimeter is measured to be 29.5 degreesC. (Given: density of water= 1.00 g/ml, specific heat of water= 4.184 J. G. -1 degrees C -1.

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Answer 1

principle of calorimeter is : heat gain = heat loss

here we can see that metal is loosing its temperature and transfer its heat to the water

so now

heat gained by water = heat loss by water

density of water =1 , volume = 75 ml , density = 1g/ml

so mass = density * volume

mass = 1*75=75 g

intial temp of water = 21.2 C , final temperature = 29.5 C change in temperature = 8.3 C

heat gained by water = mass * specific heat capacity * change in temperature

=75 * 4.184 * 8.3= 2604.54 J

now for metal = mass = 29.458g , intial temp =98.9 C , final temperature = 29.5 ,change in temperature =69.4 C

heat loss by metal = 29.458* c * 69.4 (here c is specific heat capacity)

now according to calorimetry both are equal

2604.54 = 29.458*c*69.4

c=(2604.54)/(29.458*69.4)

c=1.2739 J G-1 C-1

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