Question

Adding Ice to Water

An insulated beaker with negligible mass contains liquid water with a mass of 0.200kg and a temperature of 61.3?C .

Part A

How much ice at a temperature of -24.7?C must be dropped into the water so that the final temperature of the system will be 27.0?C ?

Take the specific heat of liquid water to be 4190J/kg?K , the specific heat of ice to be 2100J/kg?K , and the heat of fusion for water to be 3.34

Answer 1

heat gained by ice = heat gained by water....

(mi*Lf)+(mi*Si*24.7)+(mi*Sw*27) = mw*Sw*(61.3-27)....

mi(3.34*10^5)+(2100*24.7)+(4190*27) = 0.2*4190*(61.3-27)....

mi = 28743.4/((3.34*10^5)+(2100*24.7)+(4190*27)) =0.0576 kg

Answer 2

1=water
2=ice
at equilibrium absorbed heat = released heat
absorbed heat is
Qa=m2c2(273-T2)+m2k+m2c1(Te-273)
released heat is
Qr=m1c1(T-Te)
then
m2c21(T2-273)+m2k+m2c22(273-T2)=m1c1(T1...
and
m2=m1c1(T1-Te) / [c2(273-T2)+k+c1(Te-273)]
m2=0,2*4190(334,5-311,8) / [2100(273-261,6)+334x10^3+4190(311.8-273...
m2=36.5 g