300g of water at 25°c is added to 100g of ice at 0°c. the final temperature of the mixture is
Answers
Answer : The final temperature of the mixture is, 294.40 K
Solution :
As we know that,
.................(1)
where,
= mass of water = 300 g = 0.3 kg (1 kg = 1000 g)
= mass of ice = 100 g = 0.1 kg
= final temperature
= temperature of water =
= temperature of ice =
c = specific heat of water = 4185.5 J/kg K
c = specific heat of ice = 2108 J/kg K
Now put all the given values in equation (1), we get
Therefore, the final temperature of the mixture is, 294.40 K
Qs:-300 of water at 25°c is added to 100 of ice at 0°c.The final temperature of the mixture is ?
Ans:- Hot body(Water)
mass of water =300g
temparature= 25°c
Change in temperature =(25-x)
Let us suppose 'x' is the final temperature
Cold body(Ice)
mass of cold body = 100g
Temparature of ice given as 0°c
Latent heat of ice = 80 cal/g
Water at 25°c converts into water at x°c
Then heat absorbed = mc(25-x)
Ice at 0°c converts into water at 0°c and water at 0°c coverts into water at x°c
Heat gained = mL+mc (x-0)
According to principle of calorimetry
Heat absorbed = Heat gained
mc(25-x)=mL+mc (x-0)
300 (1)(25-x)=100×80+100 (0.47)(x-0)
7500-300x=8000+47x
-500=347x
x= -1.44
If the final temperature 'x' is negative we need to assume the final temperature as 0°c
-Student of
*PhysicsWallahAlakhpandey*