A 35.0 g ice cube and 170. g of liquid water are in an insulating container of n

Question

A 35.0 g ice cube and 170. g of liquid water are in an insulating container of negligible heat capacity. A small quantity of molten lead at its melting point of 327.5 degree C is poured into the cup. The lead solidifies and the system comes to thermal equilibrium at 8.20 degree C. How much lead was added to the cup? Lead: melting point = 327.5 degree C, heat of fusion = 6.8 cal/g, specific heat of solid = 0.031 cal/g degree C Water: heat of fusion = 80.0 cal/g, specific heat of liquid = 1.00 cal/g degree C, specific heat of solid = 0.50 cal/g degree C

Explanation / Answer

energy absorbed by ice = (35 x 80) + (35 x 1 x (8.20- 0))

= 3087 cal

(first ice changes its state to water at 0 deg C the its temp increases to 8.20 deg C)

energy absorbed by water = (170 x 1 x 8.20)
= 1394 cal

energy released by lead = (m x 6.8) + (m x 0.031 x (327.5 - 8.20))

= 6.8m + 9.90m
= 16.7m

from energy conservation,

energy absorbed = energy released

3087 + 1394 = 16.7m

m = 268.32 grams

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