Calculate the specific heat of a metal from the following experimental data. 75.
ID: 591816 • Letter: C
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. Show calculations in detail. 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. Show calculations in detail. Show calculations in detail.Explanation / Answer
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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