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Question

A shot of espresso has an average temperature after extraction of 82.2 °C. How many grams of ice at -18 °C are needed to cool the espresso to 0 °C. You can assume that the espresso will have a specific heat capacity and density equivalent to water. (Hint: ΔH_fus=6.01 kJ/mol, c_ice=2.09 J/g°C, 1 shot = 30.0 mL)

Answer 1

The amount of energy contained in an object is given by the relationship:

heat = mcΔT

where heat is the energy expressed in a unit such as Joules (for metric), BTU (for English), or other similar units for energy. Since the information is provided as metric, with Joules, we'll use Joules as the unit for heat. m is the mass (or moles) and c is the specific heat. Specific heat can, and is, be presented in a bewildering array of units, so be careful at this point. Energy may be in English or metric units, m may be in grams, moles, or other measure of amount, and T may be F, C, or K. We are given a value of c for ice fusion in units of kj/mol, but for water in J/g°C. Let''s work only in metric (Joules and grams) and be aware that we will need to work with both grams and moles. This is a good example of how keeping units in a calculation can prevent errors. All should cancel to yield just the unit desired.

ΔT=Tf−Ti, where Tf is the final temperature and Ti is the initial temperature

Think of the calculation as a 3-step process:

1) Determine how much heat needs to be removed from the expresso to bring 30.0 ml of expresso from 82.2°C to 0°C. This is the amount of heat that needs to be absorbed by the ice.

2) The next step is tricky, since the ice goes through a phase change as it warms. Furthermore, it starts at -18°C. This means we need to determine the amount of heat absorbed in tow separate steps:

1) Heat absorbed as the ice goes from -18°C to 0°C

2) Heat absorbed as it makes the phase change from solid to liquid (heat of fusion) at 0°C.

Note that the specific heat of water is 4,18 J/g°C, and we are told it is the same for the expresso.

1) Determine how much heat needs to be removed from the expresso to bring 30.0 ml of expresso from 82.2°C to 0°C.

heat = mcΔT

The specific heat of the expresso is the same as water: 4,18 J/g°C

Thuis unit contains grams and °C, so the m in the equation should be in grams and the temperature change in °C. One shot of expresso is 30.0 ml. At a water density of 1g/1ml, this is 30.0 grams of expresso. The temperature change is (0 - 82.24°C) or - 82.24°C

Heat(J) = (30.0g)(4,18 J/g°C)(- 82.24°C)

Heat(J) = -10308 Joules, or -10.308 kJ

The minus sign is an indication that heat is being removed from the system (the expresso)

Now let's determine how much ice is needed to absorb 10.308 kJ of energy. This is a tricky part, since ice will go througha transition phase as it melts. The temperature does not change (still at °C), but it does absorb a large amount of heat as it melts (ΔH_fus=6.01 kJ/mol). Note that the unit calls for moles, instead of grams. 1 mole of water contains 18 grams of water.

Let's calculate a simplified specific heat for water that starts as ice and ends as water at 0°C.

Let's use grams as the unit for specific heat (J/g°C). 1 gram of ice starts at -18°C and has a specifc heat of 2.09 J/g°C. The energy absorbed in going from -18°C to 0, without melting, is:

heat = mcΔT

heat = (1g)(2.09 J/g°C)(0°C - (- 18°C))

heat = (1g)(2.09 J/g°C)(18°C))

heat = 37.6 Joules

This is the heat required to raise the temperature of just 1 gram ice to 0°C. It doesn not include, yet, the energy of melting the ice. Let's add that:

heat = mc for fusion and vaporization steps, since the temperature does not change.

heat = (1g)(6.01kJ/mole)

Note that the units don't match. We need to express the 1 gram of water as moles of water.

1 gram water, with a molar mass of 18 g/mole, would be:

(1 gram)/(18 g/mole) = 0.0556 moles

So now we can write:

heat = (0.0556 mole)(6.01kJ/mole)

heat = 0.3334 kJ for 1 gram ice to go from ice to liquid water at 0°C.

Now let's combine these two into one specific heat for ice going to liquid water at 0°C:

1 gram of ice from -18°C to ice at 0°C = 37.6 Joules

1 gram of ice to water, both at 0°C = 333.34 Joules

1 gram ice going from -18°C to water at 0°C = 71.0 joules/g

Each gram of ice will require 71.0 Joules to go from -18°C to 0°C water.

3) Calculate the grams ice needed to absorb 10.308 kJ from the expresso.

From the first calculation, we know we'll need to remove 10.308 kJ (-10.308 kJ) from the expresso. From above we can count on 1 gram of ice at -18°C to absorb 71.0 joules for a final 0°C state.

(10308 J)/(71.0 joules/g) = 145.2 grams of ice at -18°C

Answer 2

Based on the law of conservation of energy, the heat LOST by the espresso must equal the heat GAINED by the ice. The equation for heat is q = mC∆T

q = heat

m = mass

∆T = change in temperature

When there is no change in temperature (phase change) we use

q = m∆H_fus

In this problem we will set the heat lost by the espresso equal to the heat gained by the ice

heat lost by espresso = (30.0 g)(4.184 J/gº)(82.2º)

heat gained by ice will be heat to get the ice from -18º to 0º PLUS heat to melt the ice at 0º

The ∆H_fus is provided in kJ/mol, which I'll convert to J/g using molar mass 18 g /mol for H2O

heat gained by ice = (x g)(2.09 J/gº)(18º) + (x g)(334 J/g)

Setting these two equal and solving for mass of ice, we have ...

(30.0 g)(4.184 J/gº)(82.2º) = (x g)(2.09 J/gº)(18º) + (x g)(334 J/g)

10318 = 37.6x + 334x

10318 = 371.6x

x = 27.8 g of ice