The answer is "c" because solidification or freezing is the term used for the process in which a liquid becomes a solid.<span> Freezing is an exothermic process that also is an example of a phase transition</span>
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
Answer:
C6 H10 O5+ 6O2-----> 6CO2+5H2O+heat
Explanation:
There are 6 carbon atoms in reactants to balance you put coefficient 6.
This makes the oxygen in CO2 12 so to balance put 6 in oxygen in reactants.
There are 10 hydrogen atoms in reactants to balance you put 5 in front of H2O in products.
If u recheck: 6 C atoms in reactants, 6 C atoms in products.
10 H atoms in reactants, 10 H atoms in products.
17 O atoms in reactants, 17 O atoms in products.
Answer:
you can write plastic , paper , stone e.t.c
Explanation:
non metals means who are not being attracted by magnets
or
which does not have good conductor of heat and electricity
some others examples can be sulpher
Answer:
The concentration of cyclobutane after 875 seconds is approximately 0.000961 M
Explanation:
The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M
The final concentration of cyclobutane, [
] = 0.00119 M
The time for the reaction, t = 455 seconds
Therefore, the Rate Law for the first order reaction is presented as follows;
![\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }](https://tex.z-dn.net/?f=%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D%20%5Ctext%20%7B-k%7D%20%5Ccdot%20t%20%7D)
Therefore, we get;
![k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%7D%20%20%7B-t%20%7D)
Which gives;
k ≈ 5.088 × 10⁻⁴ s⁻¹
The concentration after 875 seconds is given as follows;
[] = [A₀]·
Therefore;
[] = 0.00150 × 
= 0.000961
The concentration of cyclobutane after 875 seconds, [] ≈ 0.000961 M
