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alexira [117]
3 years ago
9

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.°

C), cw = 4186 J/(kg.°C), Lf= 3.35 Ă— 10 5 J/kg, Lv= 2.26 Ă— 10 6 J/kg) kg
Physics
1 answer:
muminat3 years ago
4 0

m = mass of the ice added = ?

M = mass of water = 1.90 kg

c_{w} = specific heat of the water = 4186 J/(kg ⁰C)

c_{i}  = specific heat of the ice = 2000 J/(kg ⁰C)

L_{f} = latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

T_{ii}  = initial temperature of ice = 0 ⁰C

T_{wi} = 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 c_{w}  (T - T_{ii} ) + m L_{f}  = M c_{w}  (T_{wi} - 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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