How much ice (at 0°C) must be added to 1.90 kg of water at 79 °C so as to end up with all liquid at 8 °C? (ci = 2000 J/(kg.°
1 answer:
m = mass of the ice added = ?
M = mass of water = 1.90 kg
= specific heat of the water = 4186 J/(kg ⁰C)
= specific heat of the ice = 2000 J/(kg ⁰C)
= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg
= initial temperature of ice = 0 ⁰C
= initial temperature of water = 79 ⁰C
T = final equilibrium temperature = 8 ⁰C
using conservation of heat
Heat gained by ice = Heat lost by water
m (T -
) + m
= M
(
- T)
inserting the values
m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)
m = 1.53 kg
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