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baherus [9]
2 years ago
11

Explain the difference between heat and temperature. Does 1 L of water at 658°F have more, less, or the same quantity of energy

as 1 L of water at 65°C?
Chemistry
1 answer:
pogonyaev2 years ago
3 0

The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.

Heat is a type of energy that causes a person's body to feel hot or cold.

While the temperature of an object is a parameter that indicates how hot or cold the object is.

How is the temperature in degree Fahrenheit converted to degree celsius?

  • To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.

°C =( ^0F -32) * \frac{5}{9}

  • Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
  • Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-

°C = (658 F-32) *\frac{5}{9} = 347.78 \° C

  • Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.

To learn more about heat and temperature, visit:

brainly.com/question/20038450

#SPJ4

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I have added 15 L of air to a balloon at sea level (1.0 atm). If I take the
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3 0
3 years ago
A sample of the sugar d-ribose (C5H10O5) of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess
alexandr1967 [171]

Answer:

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

Explanation:

Step 1: Data given

Mass of d-ribose = 0.727 grams

The temperature rose by 0.910 K

In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is −3251 kJ mol−1, gave a temperature rise of 1.940 K.

Molar mass of benzoic acid = 122.12 g/mol

Step 2: Calculate ΔU  for benzoic acid

The calorimeter is a constant-volume instrument so:

ΔU = q

ΔU = (0.825 g/ 122.12 g/mol) * (−3251 kJ /mol)

ΔU = -21.96 kJ

Step 3: Calculate ΔU  for d-ribose

c = |q| / ΔT

⇒ with ΔT = 1.940 K

c = 21.96 kJ / 1.940 K

c = 11.32 kJ /K

For d-ribose: ΔU = -cΔT

ΔU  = -11.32 kJ/K * 0.910 K

ΔU = - 10.3 kJ

Step 4: Calculate moles of d-ribose

moles ribose = 0.727 grams / 150.13 g/mol

moles ribose = 0.00484 moles

Step 5: Calculate the internal energy of combustion for d-ribose

ΔrU = ΔU / n

ΔrU  = -10.3 kJ / 0.004842 moles

ΔrU = -2127 kJ/mol

Step 6: Calculate The enthalpy of formation of d-ribose

The combustion of ribose is:

C5H10O5(s) + 5O2(g) → 5CO2(g) + 5H20(l)

Since there is no change in the number of moles of gas,  ΔrH = ΔU  

For the combustion of ribose, we consider the following reactions:

5CO2(g) + 5H2O(l) → C5H10O5(s) +5O2(g)      ΔH = -2127 kJ/mol

C(s) + O2(g) → CO2(g)      ΔH = -393.5 kJ/mol

H2(g) + 1/2 O2(g) → H2O(l)    ΔH = -285.83 kJ/mol

ΔH = 2127 kJ/mol + 5(-393.5 kJ/mol) + 5(-285.83 kJ/mol)

ΔH = 2127 kJ/mol - -1967.5 kJ/mol - 1429.15 kJ/mol

ΔH =  -1269.65 kJ/mol

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

3 0
3 years ago
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