Liquid ammonia (boiling point = –33.4C at 1 atm) can be used as a refrigerant and heat transfer fluid. How much energy is needed to heat 25.0 g of NH3(l) from –65.0C to –12.0C at 1 atm?

39.449 kJ.

Explanation:

q = m * Cp * Delta T

Where,

Delta T = temperature change in K

Cp = specific heat capacity in kJ/g.K

m = mass of the substance.

Delta T =33.4°C - (-65)°C

= 31.6 °C

q = 31.6 * 4.7 * 25

= 3713 J

= 3.713 kJ.

q = n * molar heat of vapourisation

Number of moles = mass/molar mass

= 25/17

= 1.47 mol.

q = 1.47 * 23500

= 34559 J

= 34.559 kJ.

q = m * Cp * Delta T

Delta T = -12°C - (-33.4)°C

= 21.4 °C

= 25 * 2.2 * 21.4

= 1177 J

= 1.177 kJ.

Adding all the energies together:

1.177 + 34.559 + 3.713

= 39.449 kJ

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2