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21.A ball at 30°C falls in water at10°C, final temperature of eachwill be * a. 10 degree celsius b. 20 degree celsius c. 40 degree celsius​

Answer»

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



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