15 g of lead (specific heat = 0.128 J/g Celsius) at 120 Degrees Celsius is placed on a glacier of ice at 0 degrees Celsius. How

Question

3 years ago

10

much ice melts? (Lf of ice = 333 J/g)

1 answer:

Anastaziya [24] · Answer 1 · 2022-03-28T19:31:19+03:00

This can be solve using heat balance. The amount of heat the lead released is equal to the amount of heat the ice absorb.

The heat release by the leas = m1*Cp*deltaT

Where m1 is the mass of lead

Cp is the specific heat of lead

Delta is the change of temperature

Heat absorbed by the ice = m2Lf

Where m2 is the mass of ice

Lf is the latent heat of fusion of ice

M1*Cp*deltaT = m2Lf

(15g) ( 0.128 j/g C) ( 120 -0 C) = m2(333/g)

Solving for m2

<span>M2 = 0.69 kg of ice melts</span>