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Sagot :
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
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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