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Sav [38]
3 years ago
7

Gold cylinder has a mass of 75 g and a specific heat of 0.129J/G degrees Celsius it is heated to 65°C and then put in 500 g of w

ater who’s temperature is 90°C what is the final temperature of the gold water system
Chemistry
1 answer:
nadezda [96]3 years ago
3 0
<h3>Answer:</h3>

89.88° C

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of gold cylinder as 75 g
  • specific heat of gold is 0.129 J/g°C
  • Initial temperature of gold cylinder is 65°C
  • Mass of water is 500 g
  • Initial temperature of water is 90 °C

We are required to calculate the final temperature;

  • We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.
  • That is, Q = m × c × ΔT
<h3>Step 1: Calculate the quantity of heat absorbed by the Gold cylinder</h3>

Assuming the final temperature is X° C

Then; ΔT = (X-65)°C

Therefore;

Q = 75 g × 0.129 J/g°C × (X-65)°C

   = 9.675X - 628.875 Joules

<h3>Step 2: Calculate the quantity of heat released by water</h3>

Taking the final temperature as X° C

Change in temperature, ΔT = (90 - X)° C

Specific heat capacity of water is 4.184 J/g°C

Therefore;

Q = 500 g × 4.184 J/g°C × (90 - X)° C

  = 188,280 -2092X joules

<h3>Step 3: Calculate the final temperature, X°C</h3>

we know that the heat gained by gold cylinder is equal to the heat released by water.

9.675X - 628.875 Joules = 188,280 -2092X joules

2101.675 X = 188908.875

              X = 89.88° C

Thus, the final temperature is 89.88° C

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