Question

200 gram of water at 50.5 degree celsius is cooled down to 10 degree Celsius per gram of ice at zero degree celsius in it find specific heat capacity of water equals to 4.2 per gram calorie and latent heat of ice equals to 336 Joel per gram

Answer 1

Correct Question :

200g of water at 50.5°C is cooled down to 10°C by adding m g of ice at 0°C in it. Find m. Take, specific heat capacity of water = 4.2 Jg⁻¹°C⁻¹ and specific latent heat of ice = 336 Jg⁻¹.

Given :

Mass of water (Mw)= 200 g

Temperature of ice (Ti) =  0° C

Specific latent heat of ice (Li)= 336 Jg⁻¹

Specific heat capacity of water (s_w) = 4.2 Jg⁻¹°C⁻¹

To find :

How much grams of ice must be added (m)

Solution :

Heat lost by water in cooling from 50.5°C to 10°C

Q₁ = Mw × s_w × ΔT₁

Q₁ = 200 × 4.2 × (50.5 - 10)

Q₁ = 200 × 4.2 × 40.5

Q₁ = 34020 J

Heat gained by ice at 0°C to change into water at 10°C

Q₂ = m × Li + m × ΔT₂ × s_w

Q₂ = m (Li + ΔT₂ × s_w)

Q₂ = m (336 + (10 - 0) × 4.2)

Q₂ = m (336 + 10 × 4.2)

Q₂ = m (336 + 42)

Q₂ = m × 378

Heat lost = Heat gained

So, Q₁ = Q₂

200 × 4.2 × 40.5 = m × 378

m = (200 × 4.2 × 40.5) / 378

m = 34020 / 378

m = 90 g

Therefore, 90 g of ice is needed.

Answer 2

Explanation:

ml+mc∆t = mc∆t

m(l+c∆t) = mc∆t