c) Suppose you have 2 cm x 2 cm x 2 cm ice cube that is frozen at a temperature

Question

c) Suppose you have 2 cm x 2 cm x 2 cm ice cube that is frozen at a temperature of 0°C. How much energy input would be required to melt it

d) Suppose you are making yourself 0.60 liters of tea but its temperature is 80°C, which is too hot for you to drink. In order to cool its temperature, you immerse three ice cubes of the same dimension as the one in c) in your tea. If the energy used to melt the ice came from the internal energy (i.e. temperature) of the tea, what is the new temperature of your tea (after the ice fully melts)?

e) Suppose you do not drink the all of the tea, and you leave 45 ml in the cup. How much energy would be required to fully evaporate the remaining tea? If this energy came from the air in a room surrounding the tea, how would the air temperature change, i.e. by what amount and as increase or decrease? Assume the room has dimensions of 5 m x 5 m x 3 m and the air average density is 1.2 kg/m³?

Explanation / Answer

Solution:

c) Energy required to melt the ice = mL

where m = mass of the ice , L = latent heat of ice = 3.34 x 10^5 J/kgC

mass of ice = density of ice x volume = 916.7 x (8x10^-6) = 0.0073 kg

Energy required to melt ice = mL = 0.0073 x (3.34x10^5) = 2449 j

b) Heat lost by tea = heat gained by the ice

(0.6)(4186)(80-t) = (0.0073)(2100)(t)+2449

=> 200928 - 2511.6 t = 15.33t +2449

=> 2527 t = 200928-2449= 198479

=> t = 198479 / 2527 = 78.5 C = final temperature of the tea.

e) Latent heat of vaporization of steam = 2.26x10^6 j/kg

to evaporate 45 ml of tea, energy required = Q = mLvap.

= 0.045 * 2.26 x 10^6 = 101700 J

Mass of air = m = density x volume = 1.2 x ( 5 x 5 x 3) = 90 kg

specific heat of air = 0.72 j/kg C

m cair T = -101700

=> T = (101700) / (1011)(90) = - 1.12 C

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