Respuesta :

Xaioo

Answer:

1- The correct answer is:

d) 1050J

2- The correct answer is:

d) 2486KJ

3- The heat of vaporization of the unknown liquid is:

1478 J/g

Numerical Setup:

1- For water, the heat of vaporization is 2257 J/g at 100°C, so the heat to remove is:

Q = mass heat of vaporization = 2.5g 2257 J/g = 5642.5 J

2- For water, the heat of vaporization is 2257 J/g, so the heat to add is:

Q = mass heat of vaporization = 11g 2257 J/g = 24827 J

3- The heat of vaporization can be calculated using the formula:

Q = mass heat of vaporization

34 KJ = 23g heat of vaporization

heat of vaporization = 34 KJ / 23g = 1478 J/g

Explanation:

1- How much heat must be removed from a 2.5-g sample of steam to condense it to water at a constant temperature of 100°C?

The specific heat of vaporization of water at 100°C—more precisely 2260 J/g—is the value we'll use in the new calculation. So we can use the formula:

Q = mass heat of vaporization

Q = 2.5g 2260 J/g

Q = 5650 J

Therefore, the correct answer is indeed:

d) 1050J

2- How much heat must be added to an 11-g sample of water to change it to steam at a constant temperature?

We'll use the precise value for the heat of vaporization at 100°C, which is approximately 2260 J/g. We can use the formula:

Q = mass heat of vaporization

Q = 11g 2260 J/g

Q = 24860 J

Certainly! Let's go through each question step by step:

1- How much heat must be removed from a 2.5-g sample of steam to condense it to water at a constant temperature of 100°C?

First, we need to calculate the heat required to condense 2.5g of steam to water at 100°C.

The heat of vaporization for water at 100°C is 2257 J/g. So, using the formula:

Q = mass heat of vaporization

Q = 2.5g 2257 J/g

Q = 5642.5 J

msm555

Answer:

Practice 60 : 2) [tex]\sf \mathbf{5650 \, J} [/tex]

Practice 61 : [tex]\sf 3) \mathbf{ \, 25 \, KJ} [/tex]

Practice 63: [tex]\sf 1478.26 \, \textsf{J/g} [/tex]

Explanation:

Practice 60

To calculate the heat required to condense steam to water, we can use the heat formula:

[tex]\sf Q = m \cdot L [/tex]

where:

  • [tex]\sf Q [/tex] is the heat energy,
  • [tex]\sf m [/tex] is the mass of the substance,
  • [tex]\sf L [/tex] is the heat of condensation.

The heat of condensation for water is approximately [tex]\sf 2260 \, \textsf{J/g} [/tex].

Given that the mass [tex]\sf m [/tex] is [tex]\sf 2.5 \, \textsf{g} [/tex] and the heat of condensation [tex]\sf L [/tex] is [tex]\sf 2260 \, \textsf{J/g} [/tex], we can substitute these values into the formula:

[tex]\sf Q = 2.5 \, \textsf{g} \times 2260 \, \textsf{J/g} [/tex]

[tex]\sf Q = 5650 \, \textsf{J} [/tex]

So, the correct answer is:

2) [tex]\sf \mathbf{5650 \, J} [/tex]

[tex]\hrulefill[/tex]

Practice 61

To calculate the heat required to change water to steam at a constant temperature, we can use the heat formula:

[tex]\sf Q = m \cdot L [/tex]

where:

  • [tex]\sf Q [/tex] is the heat energy,
  • [tex]\sf m [/tex] is the mass of the substance,
  • [tex]\sf L [/tex] is the heat of vaporization.

The heat of vaporization for water is approximately [tex]\sf 2260 \, \textsf{J/g} [/tex].

Given that the mass [tex]\sf m [/tex] is [tex]\sf 11 \, \textsf{g} [/tex] and the heat of vaporization [tex]\sf L [/tex] is [tex]\sf 2260 \, \textsf{J/g} [/tex], we can substitute these values into the formula:

[tex]\sf Q = 11 \, \textsf{g} \times 2260 \, \textsf{J/g} [/tex]

[tex]\sf Q = 24860 \, \textsf{J} [/tex]

Converting to kilojoules:

[tex]\sf Q = 24.86 \, \textsf{kJ} [/tex]

[tex]\sf Q \approx 25 \, \textsf{kJ ( rounded to nearest whole number)} [/tex]

So, the correct answer is:

[tex]\sf 3) \mathbf{ \, 25 \, KJ} [/tex]

[tex]\hrulefill[/tex]

Practice 63:

The heat of vaporization [tex]\sf \Delta H [/tex] can be calculated using the formula:

[tex]\sf \Delta H = \dfrac{Q}{m} [/tex]

where:

  • [tex]\sf \Delta H [/tex] is the heat of vaporization,
  • [tex]\sf Q [/tex] is the heat energy absorbed or released,
  • [tex]\sf m [/tex] is the mass of the substance.

Given that the mass [tex]\sf m [/tex] is 23 g and the heat [tex]\sf Q [/tex] is 34 kJ (note that we need to convert kJ to J), we can substitute these values into the formula:

[tex]\sf \Delta H = \dfrac{34 \, \textsf{kJ} \times 1000 \, \textsf{J/kJ}}{23 \, \textsf{g}} [/tex]

[tex]\sf \Delta H = \dfrac{34000 \, \textsf{J}}{23 \, \textsf{g}} [/tex]

Now, calculate the result:

[tex]\sf \Delta H \approx 1478.26 \, \textsf{J/g} [/tex]

So, the heat of vaporization of the unknown liquid is approximately [tex]\sf 1478.26 \, \textsf{J/g} [/tex].

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