Question

100 g of ice at -20°C are mixed with 250 g of water at 20°C in an insulated calorimeter. What is the final temperature of the system? How many grams of liquid water and how many grams of ice will you find after the system equilibrates? find T in degrees C; m of solid (in grams); m of liquid (in grams)

T=____

m_s=____

m_liq=____

Answer 1

specific capicity of ice(c1) = 0.53 cal/(g*k)
let heat required in ice converted to -20 oC to 0 oC be H1
H1 = m*c1*delta T
m = mass of ice
= 100 g
H1 = (100*0.53*20) cal
= 1060 cal

specific capicity of water(c2) = 1 cal/(g*k)
let heat realesed in water converted to -20 oC to 0 oC be H2
H2 = m*c2*delta T
m = mass of water
= 250 g
H2 = (250*1*20) cal
= 5000 cal

since, H2>H1
so ice will start melting at 0 oC
let amount of ice melted be x g
and let H3 be heat required to melt x g of ice
H3 = (Lf*m)
Lf = 80 cal/g
m = mass of ice
= x g
H3 = (80*x) cal

now, applying pricipal of calirometry
H1+H3 = H2
1060+H3 = 5000
H3 = 3940
80*x = 3940
x = 49.25 g

so,
final temperature = 0 oC
amount of ice left = (100 - x) g
= (100-49.25) g
= 50.75 g
final amount of water = (250+x) g
= (250+49.25) g
= 299.25 g