Question

a metal ball of specific gravity 4.5 and specific heat 0.1 cal /gm - °C is placed on a large slab of ice at 0°C half of the ball sinks in the ice. the initial temperature of the ball is

Answer 1

80degrees celcius the right answer

Answer 2

Answer:

The initial temperature of the ball is 80°C

Explanation:

Let the specific gravity of the ball is ρ(ball)=4.5 (Given)

and we know the specific gravity of the ice is ρ(ice)=0.9

Let assume that the volume of the ball is V(ball) and the volume of the ice is V(ice)

Now as in the question the ball sinks in the ice then we can say that,

V(ball)/2=V(ice).................................(1)

Let the mass of the ball is M(ball) and mass of the ice is M(ice)

We can say that V(ball)=M(ball)/ρ(ball)

and V(ice)=M(ice)/ρ(ice)

Put the volumes in the equation 1

and we get,

1/2(M(ball)/ρ(ball))=M(ice)/ρ(ice)

M(ball)/M(ice)=2x(ρ(ball)/ρ(ice))

=>M(ball)/M(ice)=2x(4.5/0.9)=10

=>M(ball)=10xM(ice)

We know latent heat of ice is 80

specific heat is 0.1

then

m(ball)x(0.1)x ΔT= M(ice)x(latent heat of ice)

=> m(ball)x(0.1)x (T-0)= M(ice) x 80

=> T=80°c