Question

100 g of hot water at 90°c is mixed to 400 g of cold water at 10°c, the equilibrium temperature of the mixture is:

Answer 1

The water originally at 90 degrees temperature will "loose" an amount of heat equal to: Delta Q = c*200 [grams]*(90-T) [degrees].

This same amount of heat will be absorbed by the water originally at 30 degrees to raise its temperature to T. Delta Q = c*100*(T-30).



For temperatures between the freezing and boiling point of water, the heating curve is linear. Thus, we can use:

((200 * 90) + (100* 30))/300

Or: ((200 * 90)(temperature of first sample weighted by mass) + (100 * 30)(temperature of second sample weighted by mass) / (300)(total mass)

to find the final temperature.

100 can be factored out to get (2 * 90) + (1 * 30)) / 3

3 can be factored out every term: (2 * 30) + (1 * 10)

(The factoring out just makes mental math easier; if a calculator is available doing so is unnecessary.)

And then just work through the order of operations.

60 + 10

=70.

So

By calorimeter principle,

Heat given by high temp fluid=Heat taken by low temp fluid

Let, final temp of water is T

So m1*c1*(T1-T)=m2*c2*(T-T2)

Because both fluid are water so c1=c2

200*(90-T)=100*(T-30)

T=70℃

You can directly use formula

T=(m1*c1*T1+m2*c2*T2)/(m1*c1+m2c2)

You will get same result T=70℃