It should be watts. <span>A </span>watt<span> is the number of </span>joules generated per second. <span>A kilocalorie is 4,184 </span>joules<span>, the amount of energy that will heat 1 kilogram of water by 1 degree </span>Celsius<span>. </span>
The total quantity of heat evolved in converting the steam to ice is determined as -12,928.68 J.
<h3>
Heat evolved in converting the steam to ice</h3>
The total heat evolved is calculated as follows;
Q(tot) = Q1(steam to boiling point) + Q2(boiling point to ice) +Q3(freezing to -42 ⁰C)
where;
Q = = mcΔθ
where;
- m is mass, (mass of water = 18 g/mol)
- c is specific heat capacity,
- Δθ is change in temperature
Q(tot) = 2(18)(2.01)(100 - 135) + 2(18)(2.01)(0 - 100) + 2(18)(2.09)(-42 - 0)
Q(tot) = -12,928.68 J
Thus, the total quantity of heat evolved in converting the steam to ice is determined as -12,928.68 J.
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Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O%29%7D%29%2B%282%20mol%5Ctimes%5CDelta%20H_f_%7B%28H_2O%29%7D%20%29%5D-%5B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2H_4%29%7D%29%2B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O_4%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%2081.6%20kJ%2Fmol%29%2B2%20mol%5Ctimes%20-241.8%20kJ%2Fmol%29%5D-%5B%281%20mol%5Ctimes%20%2850.6%20kJ%2Fmol%29%29%2B%281%20mol%5Ctimes%20%289.16%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-380.16%20kJ)
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
Answer: Using more fossil fuels
Explanation: Burning fossil fuels releases Green House gasses into the atmosphere, such as carbon dioxide. The carbon dioxide in the atmosphere traps in heat, which causes global temperatures to increase.