Water needs to be turned into steam in a high altitude lab where the atmospheric pressure is 84.6 kpa. computte the heat energy (in calories) required to evaporate 900g of water at 15 degree c under these conditions.

558.1918 kilocalories = 558191.8 calories

Explanation:

Data provided in the question:

Atmospheric pressure = 84.6 KPa

Mass of water, m = 900 g = 0.90 kg

Temperature = 15°C

Now,

Temperature at 84.6 KPa = 94.77°C

Therefore,

Heat energy required = m(CΔT + L)

here,

C is the specific heat of the water = 4.2 KJ/kg.°C

L = Latent heat of water = 2260 KJ/kg

Thus,

Heat energy required = 0.90[ 4.2 × (94.77 - 15) + 2260 ]

= 2335.53 KJ

also,

1 KJ = 0.239  Kilocalories

Therefore,

2335.53 KJ = 0.239 × 2335.53 Kilocalories

= 558.1918 kilocalories = 558191.8 calories

b. check your rearview mirror

explanation:

answer:

i cant give me a pic

explanation: