A 100-g piece of copper is heated in a furnace to a temperature T_c. The copper

Question

A 100-g piece of copper is heated in a furnace to a temperature T_c. The copper is then inserted into a 150-g copper calorimeter containing 200 g of water. The initial temperature of the water and calorimeter is 16.0 degree C, and the temperature in equilibrium is 35.0 degree C. When the calorimeter and its contents are weighed, it was found that 1.20 g of water evaporated. The specific heat of copper is C_Cu = 390 J/(kg-K), specific heat of water is c_w = 4190 J/(kg. K)and heat of vaporization is L_v = 2256000 J/kg. What was the temperature T_c?

Explanation / Answer

Here consider -

heated specimen --> 1

calorimeter cup --> 2
water in calorimeter cup --> 3

evaporated water --> 4

Now, separating the water evaporated from the initial amount of water, we get =>
Heat gained QG = m c T + ( m - m ) c T + m c T + m L
Now, m c T = [ ( 1.2 / 1000 ) kg ] [ 4190 J / ( kg • °C ) ] ( 100 – 16 )°C
= ( 1.2 / 1000 ) [ 4190 J ] ( 84 )

= 422.352 J
Again, m L = [ ( 1.2 / 1000 ) kg ] ( 2256 kJ / kg ) = 2.707 kJ = 2712 J

m - m = ( 200 / 1000 ) kg - ( 1.2 / 1000 ) kg = 0.1988 kg
So, ( m - m ) c T = ( 0.1988 kg ) [ 4190 J / ( kg • °C ) ] ( 35 – 16 )°C
= ( 832.97 J ) ( 19 )
=> m c T = [ ( 150 / 1000 ) kg ] [ 390 J /( kg • °C ) ] ( 35 – 16 )°C
= ( 58.5 J ) ( 19 )

So,
QG = ( 58.5 J ) ( 19 ) + ( 832.97 J ) ( 19 ) + 422.352 J + 2712 J …
= [ ( 58.5 J ) + ( 832.97 J ) ] ( 19 ) + 3134.352 J
= ( 891.47 J ) ( 19 ) + 3134.352 J
= 16937.93 J + 3134.352 J = 20072.282 J

Heat loss QL = m c T = [ ( 100 / 1000 ) kg ] [ 390 J /( kg • °C ) ] ( 35 – T )°C
= ( 39 J ) ( 35 – T )
Again, by conservation of energy

QG + QL = 0

So -
20072.282 J + ( 39 J ) ( 35 – T ) = 0
=> ( 20072.282 J / 39 J ) + ( 35 – T ) = 0
=> 514.67 + 35 = T
=> T = 549.67°C

So, the temperature, Tc = 549.67°C

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