1.0 kg ice from a freezer at -20 degree C is added to a reservoir of 2.0 kg of w

Question

1.0 kg ice from a freezer at -20 degree C is added to a reservoir of 2.0 kg of water at 80 degree C. Assuming there heat lost while the mixture reaches thermal equilibrium. Specific heat of water is 4 l 86 J/(kg K). Specific heat of ice is 2090 J/(kg K). Latent heat between ice and water is 3.35 times 10^5 J/(kg ). a) How much heat is required to increase the temperature of ice to 0 degree C? b) How much heat is required to melt the ice to water at 0 degree C? c) What is the final temperature of the mixture?

Explanation / Answer

Given

mass of ice m1 = 1.0 kg, at temperature T1 = -20 0C

mass of water m2 = 2 kg , at temperature T2= 80 0C

specific heat of water is Cw = 4186 J/kg k

specific heat of ice is Ci 2090 J/kg k

latent heat of fusion of ice is, L = 3.35*10^5 J /kg k

a) Amount of heat required to increase the temperature of ice to 0C

Q1 = mc*DT

Q1 = m1*Ci*DT

Q1 = 1*2090*(0-(-20) J

Q1 = 41800 J

b) heat required to melt the ice to water at 0 C is

here it is the phase change from solid to liquid so heat required is

Q2 = m1*L

Q2 = 1.0*3.35*10^5 J

Q2 = 335 kJ

c)final temperature of the mixture is T = ?

here heat lost by water = heat gain by ice

for ice Qi = 1.0*2090(0+20)+1.0*3.35*10^5*1.0*2090*(T-0)

heat lost by water Qw = 2.0*4186*(T-80)

now Qi = Qw

1.0*2090(20)+1.0*3.35*10^5+1.0*2090*(T-0) = 2.0*4186*(80-T)

T = 28 0C = 301.15 k

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