Question

10. A 140.0-g sample of water at 25.0°C is mixed with 100.0 g of a certain metal at 100.0°C. After thermal equilibrium is established, the (final) temperature of the mixture is 29.6°C. What is the heat capacity of the metal, assuming it is constant over the temperature range concerned? (A) 0.38 l/8°C (B) 0.76 l/8°C (C) 0.96 l/8°C (D) 0.031 J/gºC (E) none of these

Answer 1

10. Ans:- Option (A) " 0.38 J/g.°C "  is the correct answer.

Explanation :-

Heat evolve by water (q) = weight of water (w) x Heat capacity of water(C) x Change in temperature (T_final-T_initial)

= 140.0 g x 4.182 J/g°C x (29.6 - 25.0)°C

= - 2693.208 J

Now,

Heat evolve by water (q) = Heat absorb by metal = 2693.208   J

Again,

Heat absorb by metal (q) = weight of metal (w) x Heat capacity of metal(C_m) x Change in temperature (T_final-T_initial)

- 2693.208 J = 100.0 g x C_m x (29.6 - 100)°C

C_m = - 2693.208 J / (100.0 g x (29.6 - 100)°C)

C_m = 0.38 J/g.°C

Hence, Option (A) is the correct answer.

----------------------------------------

12. Ans :- Option (B) " 139 g"  is the correct answer.

Explanation :-

5314 KJ of heat produced by = 2 moles (that is 2 x 58.12 g) of butane

So,

6375 KJ of heat is produced by = 2 x 58.12 g x 6375 KJ / 5314 KJ

= 139.44 g of butane

Hence, Option (B) is the correct answer.