Question

A 25 g glass tumbler contains 360 mL of water at 24°C. If two 20 g ice cubes each at a temperature of-3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room

Answer 1

mass of glass tumbler mg = 25 g

mass of ice = mi = 20 g

specific heat capacity of glass = 0.84 Jkg/k

specific heat capacity of water = 4.187 Jkg/k

specific heat capacity of ice = 2.108 Jkg/k

latent heat of fusion of water = 333J/kg

heat loose by glass = 25*0.84*(24-T) = 21(24-T)

heat loose by water = 360*4.187*(24-T) = 1507.32(24-T)

heat absorbed by the ice warming = 2*20*2.108*3 = 252.96

heat absorbed by the ice melting = 2*20*333 = 13320

heat absorbed by the melted ice warming = 2*20*4.187(T-0) = 167.48 T

total heat loss = total heat gain

21(24-T)+1507.32(24-T) = 252.96+13320+167.48T

(24-T)*1528.32 = 13572.96+167.48 T

23106.72 = 1695.8T

T = 13.6 degrees