I need the answer to part B and part C The density of liquid oxygen at its bolli

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I need the answer to part B and part C

The density of liquid oxygen at its bolling point is 1.14 kg/L and its heat of vaporization is 213 kJ/kg How much energy in joules would be absorbed by 4.0 L of liquid oxygen as it vaporized? Express your answer to two significant figures and include the appropriate units. View Available Hint(s) 9.7x103 J Previous Answers Correct Obtaining liquid oxygen is difficult and expensive. Though a liquid oxygen "air conditioner" is a neat idea, it is not a very practical one. Part B cal Water's heat of fusion is 80 cal/g, and its specific heat is Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held 1500 g of frozen water at 0 C, and the temperature of the water at the end of the ride was 32 C, how many calories of heat energy were absorbed? Express your answer to two significant figures and include the appropriate units. View Available Hint's) Value Units

Explanation / Answer

Part B

We have 1500 g of frozen water at 0°C, and at the end of the ride we have water at 32°C.

To calculate the calories of heat energy absorbed by the water, we need to take into account two steps:

First, we need to calculate the heat that 1500 g of ice absorbed to melt into water at 0°C (normal temperature of water fusion). For this, we use the heat of fusion = 80 cal/g.

q1 = heat of fusion . mass = 80 cal/g . 1500 g = 120000 cal = 120 Kcal

Then, the same amount of water (1500 g) was heated to 32°C. For this we use the water specific heat = 1.0 cal/g°C.

q2 = mass . specific heat . (Tfinal – Tinitial) = 1500 g . 1.0 cal/g°C . (32°C – 0°C) = 48000 cal = 48 Kcal

As we can see, the final units are cal because g and °C are canceled in the equation.

The total heat energy absorbed by the water is the addition of these two heats:

q = q1 + q2 = 120000 cal + 48000 cal = 168000 cal

or in Kcal

q = q1 + q2 = 120 Kcal + 48 Kcal = 168 Kcal

Part C

This part is similar to part B, we have to calculate the different heats in each step.

We have 360 g of ice and 100 g of liquid water at 0°C, and all the system is heated until is converted to water vapor at 100°C (normal boiling water temperature).

So, first we have to calculate the heat that is absorbed by ice to convert into water at 0°C using the heat of fusion = 80 cal/g.

q1 = heat of fusion . mass = 80 cal/g . 360 g = 28800 cal = 28.8 Kcal

Be careful! We only use the mass from the ice!

Second, we calculate the heat that is absorbed by all the liquid water (360 g + 100 g) to heat from 0°C to 100°C (normal water boiling temperature). For this we use the water specific heat = 1.0 cal/g°C.

q2 = mass . specific heat . (Tfinal – Tinitial) = 460 g . 1.0 cal/g°C . (100°C – 0°C) = 46000 cal = 46 Kcal

Be careful! We add 360 g of water from melting ice + 100 g of the liquid water previously added to the canister, in total 460 g of liquid water.

Finally, all the liquid water is boiled / vaporized. To calculate the heat in this process, we use the heat of vaporization = 540 cal/g.

q3 = heat of vaporization . mass = 540 cal/g . 460 g = 248400 cal = 248.4 Kcal

The total heat energy absorbed by the water is the addition of these three heats:

q = q1 + q2 + q3 = 28800 cal + 46000 cal + 248400 cal = 323200 cal

or in Kcal

q = q1 + q2 + q3 = 28.8 Kcal + 46 Kcal + 248.4 Kcal = 323.2 Kcal

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