Question

Part A

An insulated beaker with negligible mass contains liquid water with a mass of 0.310 kgand a temperature of 67.5 ∘C .

How much ice at a temperature of -23.9 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?

When needed, use the constants found in the 'official' formula sheet. Assume that the heat capacity of liquid water does not depend on temperature and equals the value listed at 20°C.

Hints

mice=

kg

Part A

An insulated beaker with negligible mass contains liquid water with a mass of 0.310 kgand a temperature of 67.5 ∘C .

How much ice at a temperature of -23.9 ∘C must be dropped into the water so that the final temperature of the system will be 30.0 ∘C ?

When needed, use the constants found in the 'official' formula sheet. Assume that the heat capacity of liquid water does not depend on temperature and equals the value listed at 20°C.

Hints

mice=

kg

Answer 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) = (0.310)x(4190)x(67.5-30) = 48708.75 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-(-23.9))=(50190)M
Q(fusion)=Mx3.34x10^5=(3.34x10^5)M
Q(liquid)=Mx4190x(30-0)=(1.257x10^5)M

Heat lost by beaker liquid = Heat gained by ice when equilibrium is reached.
48708.75 = (50190)M + (3.34x10^5)M + (1.257x10^5)M
M=0.0955 kg