An insulated beaker with negligible mass contains liquid water with a mass of 0.305 kg and a temperature of 73.5 ∘C . Part A How much ice at a temperature of -22.0 ∘C must be dropped into the water so that the final temperature of the system will be 34.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg .
M = unknown mass
Qheat energyv= mass * specific heat * delta T
The amount of heat lost by the 0.305 kg of water in the beaker to the ice
Qbeaker liquid = 0.305 * 4190 * (73.5 - 34) = 50479 J
Amount of heat gained by the ice
Qice + Qfusion + Qliquid for the unknown mass M
Qice = M * 2100 * (0 - (-22)) = (46,200) M
Qfusion = M * 3.34 * 105 = (3.34 * 105 ) M
Qliquid = M * 4190 * (34 - 0) = (1.42 * 105) M
Heat lost by beaker liquid = Heat gained by ice when equilibrium is reached
50479 = (46,200) M + (3.34 * 105 ) M + (1.42 * 105) M
M = 0.0967 kg
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