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

Options

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

Explanation

c) If m is the mass of water added 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 + 20 xx 40)/(40 + 20) = 26.6^(@)`C
For option b, `T=(800 + 30 xx 30)/(40 + 30) = 24.28^(@)`C
For option (c), `T=(800 + 10 xx 60)/(40 + 10) = 28^(@)`C
For option (d), T= `(800 + 400)/(40+4)` = `27.27^(@)C`