Question

Suppose you have exactly 1 cup (237 g) of hot (100.0 °C) brewed tea in an insulated mug and that you add to it 2.50 × 10² g of ice initially at −18 °C. If all of the ice melts, what is the final temperature of the tea? Assume the tea has the same thermal properties as water.

Answer 1

Answer:

Final Temperature = 298.28 K (25.28°C)

Explanation:

First off, lets least out our parameters (What we were given).

Mass of Tea (Mt) = 237g

Initial temp. of tea (T1) = 100 °C + 273 = 373K (Converting  to Kelvin)

Mass of Ice (Mi) = 2.50 × 10² g = 250g

Initial temp. of ice (T1) = -18 °C + 273 = 255K (Converting  to Kelvin)

If all the ice melts

Amount of Heat required to melt the ice [H]  = Amount of heat required to raise its temperature to 0°C or 273K (Melting point) [H1] + Heat of Fusion [H2]

H1 = MiCΔT

ΔT = T2 - T1 = 255 - 273 = -18K

C = 2.09 J/gK (Specific Heat capacity of Ice)

H1 = 250 * 2.09 * (-18)

H1 = -9405 J

H2 = m·ΔHf

ΔHf = Heat of fusion of ice = 334 J/g

H2 = 250 * 334

H2 = 83500 J

Amount of Heat required to melt the ice [H] = H1 +  H2

H = -9405 + 83500 = 74095 J

This means the heat was able to supply 74095 J to the tea.

Final temperature is calculated from:

H = MtCΔT

ΔT = H/MtC

ΔT = 74095/(237 * 4.184)

ΔT = 74095 / 991.61

ΔT = 74.72 K

With a temperature difference of 74. 72 K

Final temperature  = Initial temperature - 74.72K

T2 = 373 - 74.74

T2 = 298.28 K (25.28°C)