Question

1.An insulated beaker with negligible mass contains liquid water with a mass of 0.345 kg and a temperature of 73.3 ∘C .How much ice at a temperature of -21.6 ∘C must be dropped into the water so that the final temperature of the system will be 26.0 ∘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 3.34×10⁵J/kg .

2.If a person has a dangerously high fever, submerging her in ice water is a bad idea, but an ice pack can help to quickly bring her body temperature down. How many grams of ice at 0 ∘ C will be melted in bringing down a 72 kg patient's fever from 40 ∘ C to 39 ∘ C ? Express your answer using two significant figures.

Answer 1

1)

The unknown mass of ice is M.

Q(heat energy)=(mass)(specific heat)(delta T)

The amount of heat lost by the 0.300 kg of water in the beaker to the ice is easily calculated:

Q(beaker liquid) = (0.345)x(4190)x(73.3-26) = 68374.5 joules

The amount of heat gained by the ice will be:

Q(ice) + Q(fusion) + Q(liquid) for the unknown mass M

Q(ice)=Mx2100x(0-(-21.6))=(45360)M

Q(fusion)=Mx3.34x10^5=(3.34x10^5)M

Q(liquid)=Mx4190x(26-0)=(1.09x10^5)M

Heat lost by beaker liquid = Heat gained by ice when equilibrium is reached.

68375.5 = (45360)M + (3.34x10^5)M + (1.09x10^5)M

M= 0.14 kg