toryahna
contestada

If you have 0.341 m3 of water at 25.0 ∘C in an insulated container and add 0.127 m3 of water at 95.0 ∘C, what is the final temperature f of the mixture? Use 1000 kg/m3 as the density of water at any temperature.

Respuesta :

Answer:

[tex]T_F=44.0\°C[/tex]

Explanation:

Hello!

In this case, since this calorimetry problem is about the mixing of two volumes of water, we can infer that the heat lost by the hot water is gained by the cold water so we can write:

[tex]Q_{W,hot}=-Q_{W,cold}[/tex]

Which in terms of mass, specific heat and temperature change is:

[tex]m_{W,hot}C_{W,hot}(T_F-T_{W,hot})=-m_{W,cold}C_{W,cold}(T_F-T_{W,cold})[/tex]

Clearly, the specific heat is the same for both waters and the masses, considering the given densities are:

[tex]m_{W,cold}=0.341m^3*\frac{1000kg}{1m^3} =341kg\\\\m_{W,hot}=0.127m^3*\frac{1000kg}{1m^3} =127kg\\[/tex]

Thus, we can solve for the the final temperature as shown below:

[tex]T_F=\frac{m_{W,hot}T_{W,hot}+m_{W,cold}T_{W,cold}}{m_{W,hot}+m_{W,cold}}[/tex]

And by plugging in the values we obtain:

[tex]T_F=\frac{341kg*25.0\°C+127kg*95.0\°C}{341kg+127kg}\\\\T_F=44.0\°C[/tex]

Best regards!

The final temperature of the mixture of hot and cold water at the given masses is 44 ⁰C.

The given parameters;

  • volume of the cold water, V = 0.341 m³
  • initial temperature of the wave, = 25⁰C
  • volume of water added hot water, = 0.127 m³
  • final temperature of water, = 95⁰ C

The mass of the cold water and the hot water are calculated as follows;

[tex]mass = density \times volume\\\\m_{cold} = 1000 \times 0.341 = 341 \ kg\\\\m_{hot} = 1000 \times 0.127 = 127 \ kg[/tex]

Apply the principle of conservation of energy to determine the final temperature of the mixture;

heat lost by the hot water = heat absorbed by the cold water

[tex]341 \times 4200 \times (t - 25) = 127 \times 4200 \times (95 - t)[/tex]

where;

  • t is the final temperature of the mixture

[tex]341(t- 25) = 127(95-t)\\\\341t - 8525 =12065 - 127t\\\\468t = 20590\\\\t = \frac{20590}{468} \\\\t = 44 \ ^0C[/tex]

Thus, the final temperature of the mixture of hot and cold water at the given masses is 44 ⁰C.

Learn more here:https://brainly.com/question/14854725

ACCESS MORE