Question

If specific heat capacity of water 4 200 J/kg 0C, heat of fusion of ice 336 000 J/kg, so specific heat capacity of ice is?
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Answer 1

Explanation:

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Question

A piece of ice of mass 40 g is added to 200 g of water at 50

o

C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =4200Jkg

−1

K

−1

and specific latent heat of fusion of ice =336×10

3

Jkg

−1

A

28.67

o

C

B

21.67

o

C

C

28.33

o

C

D

21.33

o

C

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Solution

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Correct option is C)

For water, m

1

=200,c

1

=4200,t

1

=50

o

C

For Ice, m

2

=40,c

2

=336000

Heat required to melt ice is mass×Latent heat of fusion=40×10

−3

×336×10

3

=13440J.

Let final temperature be T,

For water, m=200×10

−3

,c=4200,t=50

So energy supplied by water is 4200×200×10

−3

×(50−T)

After ice is melted to water, m=40×10

−3

,c=4200,t=0

So energy absorbed is 4200×40×10

−3

×T

Balancing energy,

4200×200×10

−3

×(50−T)=13440+4200×40×10

−3

×T

Solving above, T=28.33

o

C