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MAVERICK [17]
3 years ago
15

What is the density of the black hole

Chemistry
2 answers:
Karolina [17]3 years ago
7 0

the density of a black hole is 22.61 g/cm3 i think

Mila [183]3 years ago
6 0
With a density of 22.61 g/cm^3
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Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of h
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Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
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The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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