Take 100ml of water at 90 degrees Celsius and 200ml of water at 60 degrees Celsius and mix the two .what is the temperature of the mixture?
a- 70 degrees Celsius
b-7 degrees Celsius
c-80 degrees Celsius
d- 65 degrees Celsius
Answers
The temperature of the mixture is 70 degrees Celsius.
- We know that,
Heat gained by one = heat lost by the other
- Now, let us consider final temperature to be T.
- Therefore,
0.1 x (T-90) x const = 0.2 x (60-T) x const
T(0.1) - 9 = 12 - 0.2(T)
(0.3)T = 21
T = 21/0.3
= 70
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