Which one of the following would reaise the temperature of 40 g of water at `20^(@)`C most when mixed with?

Options

A. 40 g of water at `20^(@)C`
B. 30 g of water at `30^(@)C`
C. 10 g of water at `60^(@)C`
D. 4 g of water `100^(@)C`

Answer

C

Explanation

If m is the mass of water at `theta^(@)C` , then resultant temperature T is given as,
Heat lost = Heat gained
`rArr" " m(theta-T)=40(T-20)`
`rArr" " T=(800+m.theta)/(40+m)`
So ,for option (a) , `T=(800+20xx40)/(40+20)=26.6^(@)C`
For option (b), `T=(800+30xx30)/(40+30)=24.28^(@)C`
For option (c) ,`T=(800+10xx60)/(40+10)=28^(@)C`
for option (d), `T=(800+400)/(40+4)=27.27^(@)C`