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Suppose 550 g of water at 32 degrees * C is poured into a 210 - g aluminum can with an initial temperature of 15 degrees * C . Find the final temperature of the system, assuming that no thermal energy is exchanged with the surroundings.

Sagot :

janmilczek

Answer:

about 30.714degC

Explanation:

Assuming

specific heat of water at  4200 J/kg * K

specific heat of aluminum 900 J/kg * K

(actual values to higher precision differ, and also change slightly with temperature).

To raise the aluminum's temp by 1K we need 900J * (210g / 1kg) = 189J

189J given by the water decreases its temp by 1K * (189J / 4200J) * (1kg / 550g) = 0.0(81)K.

So every K the temp of aluminum rises, 0.0(81)K falls the temp of the water.

At a difference of (32degC - 15degC) = 17K, we infer the temp will balance out after 17K / (1K + 0.0(81)K) "steps", so about 15.714.

The final temperature will be around 30.714degC

onyebuchinnaji

The final temperature of the water-aluminum mixture is 30.7 ⁰C.

The given parameters;

  • mass of the water, = 550 g
  • initial temperature of the water, = 32⁰C
  • mass of the aluminum, 210 g
  • initial temperature of the aluminum = 15⁰ C
  • specific heat capacity of water = 4.184 J/g⁰C
  • specific heat capacity of aluminum = 0.9 J/g⁰C

The final temperature of the mixture is determined by applying the principle of conservation of energy;

Heat lost by the water = Heat gained by the aluminum

[tex]mc_w \Delta t = mc_{AL} \Delta t\\\\550 \times 4.184 \times (32 - t) = 210\times 0.9 \times (t - 15)\\\\73638.4 - 2301.2t = 189t - 2835\\\\76473.4 = 2490.2t\\\\t =\frac{76473.4}{2490.2} \\\\t = 30.7 \ ^0C[/tex]

Thus, the final temperature of the water-aluminum mixture is 30.7 ⁰C.

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

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