Question

600g of water at 100c is mixed with 300g at 20°c . find the temperature of the mixture. (specific heat capacity=4.2J/g*°C​

Answer 1

Answer:

Mass of hot water, m

1

=500g

Mass of cold water, m

2

=300g

Temp. of hot water, t

1

=100

o

C

Temp. of cold water, t

2

=30

o

C

Sp. heat of water, C=4.2Jg

−1

o

C

−1

Let temp. of mixture be t

o

C. Then, Heat gained by cold water

=m

2

×C×(t−t

2

)

According to the principle of calorimetry, Heat lost = Heat gained

500×4.2×(100−t)

=300×4.2×(t−30)

⇒5(100−t)=3(t−30)

⇒−3t−5t=−90−500

⇒−8t=−590

⇒t=

8

590

=73.8

o

C

So, the final temperature of the mixture is 73.8

o

C.