A 400-g block of iron at 400°C is dropped into a calorimeter (of negligible heat capacity) containing 60 g of water at 30°C. How
2 answers:
Answer:
<em>The mass of steam produced would be 21.95 g</em>
Explanation:
The heat that the iron block would lose will be gained by water, this is expressed thus;
Heat loss by Iron = Heat gain by water;
Given that;
mass of block = 400 g / 1000 = 0.4 kg
Initial temp of iron = 400°C
mass of water = 60 g / 1000 = 0.06 kg
heat vaporization of water L (constant) = 22.6 x J/Kg
average specific heat of the iron over this temperature range is = 560 J/Kg K.
Specific heat capacity of water = 4186 J / kg K
heat lost by iron = heat gain by water .......................1
ΔT is the change in temperature.
m is the mass of steam during vaporization
Heat loss by iron = x x ΔT
= 0.4 x 560 x (400-100)
= 67200 J
Heat gained by water = (ΔT ) + m L
expression 1 would be;
67200 = 17581.2 + m x (22.6 x )
Making m the subject formula we have;
m = ( 67200-17581.2 ) / ( 22.6 x )
m = 0.0219 kg x 1000 = 21.95 g
Therefore the amount of steam produced would be 21.95 g
Answer:
Amount of steam produced = m' = 16.1 grams
Explanation:
Mass of the block of iron = M = 400 g
Initial temperature of iron = = 400 °C
Mass of water = m = 60 g
Initial temperature of water = = 30 °C
Specific heat of iron = 450 J/kgC
Specific heat of water = 4186 J/kg C
Latent heat of vaporization of water = L =
Heat lost by iron = Heat gained by water + Heat of vaporization.
(400)(450)(400 - 100)= (60)(4186)(100 - 30) + m' (
⇒ 5.4 × 10⁷ = 1.7581 × 10⁷ + 22.6 × 10⁵ m'
⇒ m'=mass of steam produced =3.642 × 10⁷ / 22.6 × 10⁵ kg
= 16.1 grams
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