Question

If you have 410.0 g of water at 25.00 °C in an insulated vessel and add 120.0 g of water at 95.00 °C, what is the final temperature of the mixture?

Answer 1

Given :

Mass of hot water = 120.0 g , Initial T = 95.00 deg C

Mass of cold water = 410.0 g , Initial T = 25.00 deg C

We use following formula to calculate final T

q = m x C x Delta T

here q is heat Absorbed , C is specific heat , Delta T is change in T

when we add hot water to cold water then heat is transferred from hot cold and at equilibrium both will have same T .

That is the final temperature we are interested in.

Amount of heat given by hot water = amount of heat absorbed by cold water

Lets show this by using mathematical equation

-q ( hot) = q (cold)

Here minus sign indicates that the heat loss of hot water and heat gain of cold water.

Lets plug all the values from given data.

-120.0 g x C x (Tf- 95.0 )deg C = 410.0 g x C x (Tf-25.00) deg C

C is common since both side there is water.

-120.0 g x (Tf- 95.0 )deg C = 410.0 g (Tf-25.00) deg C

(Tf- 95.0 )deg C = - 3.42 (Tf-25.00) deg C

(Tf- 95.0 )deg C = - 3.42Tf + 85.42 deg

Tf +3.42 Tf = (85.42 + 95.00) deg C

4.42 Tf = 180.42 deg C

Tf = 40.82 deg C

So the final temperature of the water = 40.8 deg C