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

Answer:

The mass of steam produced would be 21.95 g

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

Answer 2

Answer:

Amount of steam produced = m' = 16.1 grams

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 = $22.6 \times10^{5 } J/kg$

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

(400)(450)(400 - 100)= (60)(4186)(100 - 30) + m' ($22.6 \times 10^{5 }$

⇒ 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