A calorimeter contains 0.2 kg of water at 30 ° C , in which 0.1 kg of water at 60 ° C is added. The mixture is well stirred and the resulting temperature is found to be 35 ° C . The thermal capacity of the calorimeter is

Question

A calorimeter contains 0.2 kg of water at 30 ° C , in which 0.1 kg of water at 60 ° C is added. The mixture is well stirred and the resulting temperature is found to be 35 ° C . The thermal capacity of the calorimeter is (A) 6300JK- 1 (B) 1260JK- 1 (C) 4200JK- 1 (D) 3200JK- 1

Tardigrade Admin · Accepted Answer

Let X be the thermal capacity of the calorimeter and specific heat of water = 4200Jkg- 1K- 1 . Heat lost by 0.1kg of water = Heat gained by water in calorimeter + Heat gained by calorimeter ⇒ 0.1× 4200× (.60-35.) =0.2× 4200× (.35-30.)+X(.35-30.) 10500=4200+5X ⇒ X=1260JK- 1

Q. A calorimeter contains $0.2 k g$ of water at $30^{\circ} C$ , in which $0.1 k g$ of water at $60^{\circ} C$ is added. The mixture is well stirred and the resulting temperature is found to be $35^{\circ} C$ . The thermal capacity of the calorimeter is

$6300 J K^{- 1}$

$1260 J K^{- 1}$

$4200 J K^{- 1}$

$3200 J K^{- 1}$

Q. A calorimeter contains 0.2kg of water at 30∘C , in which 0.1kg of water at 60∘C is added. The mixture is well stirred and the resulting temperature is found to be 35∘C . The thermal capacity of the calorimeter is

6300JK−1

1260JK−1

4200JK−1

3200JK−1

Let X be the thermal capacity of the calorimeter and specific heat of water = 4200Jkg−1K−1 . Heat lost by 0.1kg of water = Heat gained by water in calorimeter + Heat gained by calorimeter ⇒ 0.1×4200×(60−35) =0.2×4200×(35−30)+X(35−30) 10500=4200+5X ⇒ X=1260JK−1

Q. A calorimeter contains $0.2 k g$ of water at $30^{\circ} C$ , in which $0.1 k g$ of water at $60^{\circ} C$ is added. The mixture is well stirred and the resulting temperature is found to be $35^{\circ} C$ . The thermal capacity of the calorimeter is

$6300 J K^{- 1}$

$1260 J K^{- 1}$

$4200 J K^{- 1}$

$3200 J K^{- 1}$