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Grade 12th passThermal Physics

22 gm. of CO2 at 270 C is mixed with 16 gm. of O2 at 37°C. The temperature of the mixture is:-

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Profile image of Nikhila Thakur
9 Years agoGrade 12th pass
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1 Answer

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ApprovedApproved Tutor Answer1 Year ago

To determine the temperature of the mixture when 22 grams of CO2 at 27°C is mixed with 16 grams of O2 at 37°C, we can use the principle of conservation of energy, specifically the concept of heat transfer. The heat lost by the warmer gas will equal the heat gained by the cooler gas until thermal equilibrium is reached.

Understanding the Components

First, let's identify the specific heat capacities of the gases involved. For our calculations, we will use the following approximate values:

  • Specific heat capacity of CO2: 0.844 J/g°C
  • Specific heat capacity of O2: 0.918 J/g°C

Calculating the Heat Transfer

We can set up the equation based on the heat gained and lost:

Let T be the final temperature of the mixture. The heat lost by O2 can be expressed as:

Q_O2 = m_O2 * c_O2 * (T_initial_O2 - T)

And the heat gained by CO2 can be expressed as:

Q_CO2 = m_CO2 * c_CO2 * (T - T_initial_CO2)

Since the heat lost by O2 equals the heat gained by CO2, we can set these two equations equal to each other:

m_O2 * c_O2 * (T_initial_O2 - T) = m_CO2 * c_CO2 * (T - T_initial_CO2)

Plugging in the Values

Now, substituting the known values:

  • m_O2 = 16 g
  • c_O2 = 0.918 J/g°C
  • T_initial_O2 = 37°C
  • m_CO2 = 22 g
  • c_CO2 = 0.844 J/g°C
  • T_initial_CO2 = 27°C

Substituting these values into the equation gives:

16 g * 0.918 J/g°C * (37°C - T) = 22 g * 0.844 J/g°C * (T - 27°C)

Simplifying the Equation

Now, let's simplify this equation step by step:

14.688 (37 - T) = 18.568 (T - 27)

Expanding both sides:

543.416 - 14.688T = 18.568T - 501.336

Combining like terms:

543.416 + 501.336 = 18.568T + 14.688T

1044.752 = 33.256T

Finding the Final Temperature

Now, divide both sides by 33.256 to find T:

T = 1044.752 / 33.256 ≈ 31.4°C

Final Thoughts

The final temperature of the mixture when 22 grams of CO2 at 27°C is mixed with 16 grams of O2 at 37°C is approximately 31.4°C. This calculation illustrates how heat transfer works between two substances until they reach thermal equilibrium, showcasing the principles of thermodynamics in a practical scenario.