Question

how much heat should be extracted from 500 gram of water so that its temperature falls from 95 degree centigrade to 30 degree centigrade​

Answer 1

Explanation:

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

hope this helps you

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