Question

How many grams of liquid of specific heat 0.2 at a temperature?

Answer 1

Answer:

375gm

Explanation:

Temperature of mixture θ=  

m  

1

​  

c  

1

​  

+m  

2

​  

θ  

2

​  

 

m  

1

​  

c  

1

​  

θ  

1

​  

+m  

2

​  

c  

2

​  

θ  

2

​  

 

​  

 

⇒32=  

m  

1

​  

×0.2+100×0.5

m  

1

​  

×0.2×40+100×0.5×20

​  

⇒m  

1

​  

=375gm