Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 81.6 ∘C .

How much ice at a temperature of -23.6 ∘C must be dropped into the water so that the final temperature of the system will be 40.0 ∘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 .

mice =

Answer 1

amount of heat released when the temperature of water reduced from 81.6 to 40 ^o C

Q = mcdT

= 0.285*4190*(81.6 - 40)

= 49676.64 J

Heat required to increase the temperature of ice to 40 ^oC

= energy required to make ice from -23.6 to 0 + energy required to melt the ice + energy required to increase the temp to 40 ^oC

= m*2100*23.6 + m*3.34*10^5 + m*4190*40

heat lost by water = heat gained by ice

49676.64 = 551160 m

m = 0.0901 kg

so the mass of ice required = 90.1 gm