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16 100 g of water at 25 °C is poured into an insulating cup. 50 g of ice at 0 °C is added to the water.
The water is stirred until the temperature of the water has fallen to 0°C.
18 g of ice remains unmelted.
The specific heat capacity of water is 4.2 J /g °C.
Which value does this experiment give for the specific latent heat of fusion of ice?

Respuesta :

Answer:

Q = Q

mcT = ml

100 x 4.2 x 25 = (50-18) l

l = 328

*Hope it helps*

onyebuchinnaji

The specific latent heat of fusion of the melted ice is 328.13 J/kg.

Conservation of energy

The specific latent heat of fusion of the melted ice is determined by applying the principle of conservation of energy as shown below;

Heat lost by the water = Heat gained by the ice

McΔθ = mL

where;

  • c is specific heat capacity of water
  • L is pecific latent heat of fusion of ice
  • m is mass of melted ice = 50 g - 18 g = 32 g

(100)(4.2)(25) = (32)L

32L = 10500

L = 328.13 J/kg

Thus, the specific latent heat of fusion of the melted ice is 328.13 J/kg.

Learn more about heat capacity here: https://brainly.com/question/16559442

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