I am sure the answer is D
Answer: The reaction is exothermic. The value of q is -542 kJ.
Explanation:
Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and for the reaction comes out to be positive.
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.
Thus 
evolves heat , it is exothermic in nature. The value of q is -542kJ.
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
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