Question

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 much steam is produced?

Answer 1

Amount of steam produced = m' = 16.1 grams

Step-by-step explanation:

Mass of the block of iron = M = 400 g

Initial temperature of iron =
`t_(i)` = 400 °C

Mass of water = m = 60 g

Initial temperature of water =
`t_(w)` = 30 °C

Specific heat of iron = 450 J/kgC

Specific heat of water = 4186 J/kg C

Latent heat of vaporization of water = L =
`22.6 *10^(5 ) J/kg`

Heat lost by iron = Heat gained by water + Heat of vaporization.

(400)(450)(400 - 100)= (60)(4186)(100 - 30) + m' (
`22.6 * 10^(5 )`

⇒ 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

Answer 2

The mass of steam produced would be 21.95 g

Step-by-step 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 =
`M_(w)`
`C_(w)` (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