The heat required to vaporize 1.00 kg of ammonia is 23,400 kJ. To find the number of grams of water that could be frozen to ice by the evaporation of this amount of ammonia, we need to calculate the amount of heat released by the evaporation of 1.00 kg of ammonia. This can be done by using the equation Q = m x c x ΔT, where Q is the heat released, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature. For water, the specific heat capacity is 4.18 J/g°C.
Therefore, the amount of heat released by the evaporation of 1.00 kg of ammonia is 23,400 kJ. This amount of heat can be used to freeze 1.00 kg x 4.18 J/g°C x 0°C = 4,180 g of water at 0°C.
The heat of vaporization of ammonia is 23.4 kj/moles. How much heat is required to vaporize 1.00kg of ammonia? How many grams of water at 0 degree celcius could be frozen to ice at 0 degree celcius by evaporation of this Amount of ammonia?
I got the kj by doing the layout 1,000 g nh3 x 1/17.0 g/mol nh3 x -23.4/1 mol nh3 = 1.37 x 10 ^3 kj nh3 its the second part to find the grams is the problem for me. Please help
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