An insulated beaker with negligible mass contains liquid water with a mass of 0.295 kg and a temperature of 61.6 ∘C .
How much ice at a temperature of -12.6 ∘C must be dropped into the water so that the final temperature of the system will be 28.0 ∘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 .
*ANSWER IN KG*
amount of energy which must be removed from the water to get to 28.0 ∘C is given as
Q = 0.295 * 4190 * ( 61.6 - 28)
Q = 41531.28 J
there will be three phases for ice
take ice from initial temperature to 0 degree , then convert this to water at 0 degree and then take water from 0 degree to desired temperature
Let x be the required mass
x * 2100 * 12.6 + x * 3.34e5 + x * 4190 * 28 = 41531.28
x ( 2100 * 12.6 + 3.34e5 + 4190 * 28) = 41531.28
x = 0.087 kg
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