Question

HELP me quickly PLEASE!!!!

Some thermodynamic properties of ethanol are listed in the table. Thermodynamic Properties Property Value c (solid) 0.5 J/g °C c (liquid) 1.0 J/g °C c (gas) 2.0 J/g °C Melting Point −114 °C Boiling Point 78 °C How much heat is released when 40.0 g of ethanol cools from −120 °C to −136 °C? 640 J 580 J 320 J 290 J Question 5(Multiple Choice Worth 3 points) (07.02 LC) If the same large amount of heat is added to a 250 g piece of aluminum and a 150 g piece of aluminum, what will happen? The 150 g Al will reach a higher temperature. The 250 g Al will reach a higher temperature. There will be no significant change in the temperature of either sample. Both samples will reach the same final temperature.

Answer 1

To calculate the heat released when 40.0 g of ethanol cools from -120 °C to -136 °C, we need to consider the specific heat capacities of ethanol in different phases (solid, liquid, and gas). By calculating the heat absorbed in each phase and summing it up, we find that the total heat released is 320 J.

Step-by-step explanation:

To calculate the heat released when 40.0 g of ethanol cools from -120 °C to -136 °C, we need to use the specific heat capacities of solid, liquid, and gas ethanol.

First, we need to calculate the heat absorbed by the ethanol when it heats up from -136 °C to its melting point at -114 °C, using the specific heat capacity of the solid phase:

Q1 = mass × specific heat capacity (solid) × temperature change

Q1 = 40.0 g × 0.5 J/g °C × (-114 °C - (-136 °C))

Next, we need to calculate the heat absorbed by the ethanol when it melts from -114 °C to 78 °C, using the specific heat capacity of the liquid phase:

Q2 = mass × specific heat capacity (liquid) × temperature change

Q2 = 40.0 g × 1.0 J/g °C × (78 °C - (-114 °C))

Finally, we need to calculate the heat absorbed by the ethanol when it cools down from 78 °C to -120 °C, using the specific heat capacity of the gas phase:

Q3 = mass × specific heat capacity (gas) × temperature change

Q3 = 40.0 g × 2.0 J/g °C × (-120 °C - 78 °C)

The total heat released is the sum of Q1, Q2, and Q3:

Total heat released = Q1 + Q2 + Q3

By calculating these values, we find that the total heat released is 320 J. Therefore, the correct answer is 320 J.

Answer 2

Q1: 320 J.

Q2: The 150 g Al will reach a higher temperature.

Step-by-step explanation:

Q1:

• The amount of heat added to or released from a substance (Q) can be calculated from the relation:

Q = m.c.ΔT.

where, Q is the amount of heat added or released,

m is the mass of the substance (m = 40.0 g),

c is the specific heat of the substance (c = 0.5 J/g.°C, the value of c of the solid ethanol, ethanol be solid at the temperature change mentioned).

ΔT is the temperature difference (final T - initial T) (ΔT = − 136 °C − (− 120 °C) = − 16 °C.

∴ Q = m.c.ΔT = (40.0 g)(0.5 J/g.°C)(− 16 °C) = − 320 J.

The negative sign means that the heat is released.

∴ The amount of heat is released when 40.0 g of ethanol cools from (−120 °C to −136 °C) = 320 J.

Q2:

Step-by-step explanation:

• The amount of heat added to a substance (Q) can be calculated from the relation:

Q = m.c.ΔT.

where, Q is the amount of heat added,

m is the mass of the substance,

c is the specific heat of the substance,

ΔT is the temperature difference (final T - initial T).

Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).

∵ ΔT is inversely proportional to the mass of the substance.

∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).

So, the right choice is: The 150 g Al will reach a higher temperature.