Question

You place 20g of ice initially at -10 degrees celcius and 200g of an unknown substance initially at 90 degrees celcius together in an insukated container. You find that the final temperature of the system is 34.6 degrees celcius and note that the unknown substance did not change phase. Given that the specific heat of ice is 2.04 kJ/kg•K, the specific heat of liquid water is 4.18 kJ/kg•K, and the heat of fusion of water is 334 kJ/kg, find the specific heat of the unknown substance.

Answer 1

Heat absorbed by 20 g or 0.02 kg ice to reach at 0^o C from - 10^o C :

H₁ = 0.02 kg x 2.04 kJ/ ( kg . K ) x [ 0 - ( - 10 ) ] K = 0.408 kJ.

Heat absorbed by the ice to melt completely :

H₂ = 0.02 kg x 334 ( kJ / kg ) = 6.68 kJ.

Finally, heat absorbed by the melted ice, i.e., the 0.02 kg water to reach at 34.6^o C :

H₃ = 0.02 kg x 4.18 kJ/ ( kg . K ) x ( 34.6 - 0 ) K ~ 3 kJ.

Hence, total heat absorbed by the ice, to reach at 34.6^o C = H = H₁ + H₂ + H₃ = 10.088 kJ.

Now, if the specific heat of the unknown material be s, in kJ / ( kg . K ),

then, heat released by 200 g or 0.2 kg of this material to reach at 34.6^o C from 90^o C :

H^/ = 0.2 kg x s kJ / ( kg . K ) x ( 90 - 34.6 ) K = 11.08s kJ.

Since total heat energy is conserved in this process, we get : H^/ = H,

or, 11.08s kJ = 10.088 kJ

or, s = 10.088 / 11.08 ~ 0.91.

Hence, specific heat of the unknown substance is : 0.91 kJ / ( kg . K ), which tells that the unknown substance is aluminium.