Question

The resultant temperature of the mixture of 100 ml of water
at 90°C and 200 ml of water at 60°C is

A) 75° C
B) 70°C
C) 80° C
D) 90° C​

Answer 1

Answer:

Temperature of water ( T₁ ) = 90°C = (90 + 273) K = 363 K. Mass of water ( m₂ ) = 200 ml = 0.2 kg

Answer 2

Answer:

The final temperature is 70°C is correct.

Explanation:

The given data :-

Liquid - 1

Mass of water ( m₁ ) = 100 ml = 0.1 kg.

Temperature of water ( T₁ ) = 90°C = (90 + 273) K = 363 K.

Specific heat of water ( c₁ ) = 4.187 KJ/kgK

Liquid - 2

Mass of water ( m₂ ) = 200 ml = 0.2 kg.

Temperature of water ( T₂ ) = 60°C = (273 + 60) K = 333 K.

Specific heat of water ( c₂ ) = 4.187 KJ/kgK

After mixing the two liquid ,heat loss by a liquid is heat gained by another liquid and they attain a common final temperature.

Let the common final temperature be T.

m₁ * c₁ * ( 363 - T ) =  m₂ * c₂ * ( T - 333 )

0.1 * 4.187 * ( 363 - T ) = 0.2 * 4.187 * ( T - 333 )

363 - T = 2T  - 666

3T = 1029

T = 343 K

T = 343 - 273 = 70°C

So the final temperature is 70°C..