Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Pleas HELP!!!!!! 1. Water\'s heat of fusion is 80. cal/g , and its specific heat

ID: 916088 • Letter: P

Question

Pleas HELP!!!!!!

1. Water's heat of fusion is 80. cal/g , and its specific heat is 1.0calgC . Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held 1300 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?

2. Water's heat of fusion is 80. cal/g , its specific heat is 1.0calgC, and its heat of vaporization is 540 cal/g . A canister is filled with 340 g of ice and 100. g of liquid water, both at 0 C . The canister is placed in an oven until all the H2O has boiled off and the canister is empty. How much energy in calories was absorbed?

3. High-pressure liquid chromatography (HPLC) is a method used in chemistry and biochemistry to purify chemical substances. The pressures used in this procedure range from around 500 kilopascals (500,000 Pa) to about 60,000 kPa (60,000,000 Pa). It is often convenient to know the pressure in torr. If an HPLC procedure is running at a pressure of 5.22×108 Pa , what is its running pressure in torr?

4. A sample of ideal gas is in a sealed container. The pressure of the gas is 655

torr , and the temperature is 24 C . If the temperature changes to 87 C with no change in volume or amount of gas, what is the new pressure, P2, of the gas inside the container?

5. Using the same sample of gas (

P1 = 655 torr , T1 = 24 C ), we wish to change the pressure to 6550 torr with no accompanying change in volume or amount of gas. What temperature T2, in Celsius, is needed to reach this pressure?

6. A sample of ideal gas at room temperature occupies a volume of 11.0 L at a pressure of 452 torr . If the pressure changes to 2260 torr , with no change in the temperature or moles of gas, what is the new volume, V2?

7. If the volume of the original sample in Part A (

P1 = 452 torr , V1 = 11.0 L ) changes to 66.0 L, without a change in the temperature or moles of gas molecules, what is the new pressure, P2

Explanation / Answer

1)

heat required to melt the ice = mass x heat of fusion

so

Q1 = 1300 x 80

Q1 = 104000 cal

noq

heat required to raise the temp of water at 0 C to 32 C = mx s x dT

so

Q2 = 1300 x 1 x 32

Q2 = 41600

so

total heat absorbed = 104000 + 41600

total heat absorbed = 145600

so

145.6 kcal of heat is absorbed

2)

Q1 = heat required to melt the ice

Q1 = 340 x 80

Q1 = 27200 cal

now

total mass of water = 340 + 100 = 440

now

heat required to raise the temperature to 100 C

Q2 = m x s x dT

Q2 = 440 x 1 x 100

Q2 = 44000 cal

now

heat required to vaporize

Q3 = 440 x 540

Q3 = 237600 cal

total heat = Q1 + Q2 + Q3

total heat = 27200 + 44000 + 237600

total heat = 308.8 k cal

so

308.8 kcal of energy is absorbed

3)

we know that

1 pa = 0.0075 torr

so

pressure = 5.22 x 10^8 x 0.0075

pressure = 3.915 x 10^6 torr

so

the pressure is 3.195 x 10^6 torr

4) PV = nRT

volume is constant

so

P1/T1 = P2 / T2

655 / 297 = P2 / 360

P2 = 793.94

so

new pressure is 793.94 torr

5)

volume = constant

so

P1 / T2 = P2 / T2

655 / 297 = 6550 / T2

T2 = 2970 K

T2 = 2970 - 273

T2 = 2697

so

the temp should be 2607 C

6)

P1V1 = P2V2

452 x 11 = 2260 x V2

V2 = 2.2

so

the new volume is 2.2 L


7)

P1V1 = p2 V2

452 x 11 = P2 x 66

P2 = 75.33

so

the new pressure is 75.33 torr