Question

2.) Your Australian friend has just pointed out that American football" is poorly name begun to enumerate the many ways in which Aussie rules is the sunerior sport. You get tevens by dropping some ice into her tea. You take the ice from a -20.0 °C freezer and put it into thermos that contains 325 g of tea at 85.0 °C. (You can assume that the thermos is perfectly insulating, and no heat is transferred to or from the thermos or environment.) How much ice will it take to bring the tea down to a tepid 40.0 °C?

Answer 1

Heat lost from tea, Q1 = Mt x Ct x $\Delta$ T1
Where Mt is the mass of tea, Ct is the specific heat of tea (here we assume that it is same as that of water) and $\Delta$ T1 is the change in temperature of the tea.
Q1 = 325 x 4.186 x (85 - 40)
= 61220.25 J

Consider that Mi gram of ice is taken
Heat gained by ice
Q2 = Mi x Ci x $\Delta$ T2 + Mi x Li + Mi x Cw x $\Delta$ T3
Where Ci, Cw are specific heat of ice and water, $\Delta$ T2 and $\Delta$ T3 are the change in temperature of ice and water, and Li is the latent heat of fusion of ice
Q2 = Mi [Ci x $\Delta$ T2 + Li + Cw x $\Delta$ T3]
= Mi [2.108 x (20 - 0) + 334 + 4.186 x (40-0)]
= 543.6 Mi

Q1 = Q2
61220.25 = 543.6 Mi
Mi = 61220.25 / 543.6
= 112.6 g