Question

21.A ball at 30°C falls in water at
10°C, final temperature of each
will be * a. 10 degree celsius b. 20 degree celsius c. 40 degree celsius​

Answer 1

Given : T m=130^oC h=6200 m S=126 Jkg

−1C−1

Let the mass of the lead ball be m and its latent heat of fusion be L.

Initial temperature of the lead ball T1=30 ^oC

The potential energy of the ball gets converted into heat that melts the ball.

∴ mgh=mS(Tm−T1)+mL

OR m(10)(6200)=m(126)(130−30)+mL

OR 62000=12600+L ⟹L=4.94×10

4Jkg −1