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Water of mass, m at 100 ℃ is added to 0.50 kg of water at 20 ℃ in a lagged calorimeter
of thermal capacity 105 JK
−1
. If the specific heat capacity of water is 4200 kg
−1K
−1
and
the final temperature of the mixture is 70℃, determine the value of m

Respuesta :

the value of m, (the mass of water at 100°C) is 0.875 kg.

What is heat?

Heat is a form of energy that brings about the sensation of warmth or coldness.

Heat lost by the hot water = heat gained by the cold water + heat gained by the calorimeter.

To calculate the value of m, we use the formula below

Formula

  • cm(t₁-t₃) = cm'(t₃-t₂)+C(t₃-t₂)........... Equation 1

Where:

  • c = specific heat capacity of water
  • m = mass of the hot water
  • m' = mass of the cold water
  • C = Thermal capacity of the calorimeter
  • t₁ = Initial temperature of the hot water
  • t₂ = Initial temperature of the cold water and the calorimeter
  • t₃ = Temperature of the mixture

Make m the subject of the equation

  • m = [cm'(t₃-t₂)+C(t₃-t₂)]/c(t₁-t₃).............. Equation 2

From the question,

Given:

  • m' = 0.5 kg
  • t₁ = 100 °C
  • t₂ = 20 °C
  • t₃ = 70 °C
  • C = 105 J/K
  • c = 4200 J/kg.K

Substitute these values into equation 2

  • m = [0.5×4200(70-20)+105(70-20)]/[4200(100-70)]
  • m = (105000+5250)/(12600)
  • m = 110250/12600
  • m = 0.875 kg.

Hence the value of m is 0.875 kg.

Learn more about heat here: https://brainly.com/question/13439286

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