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 water who’s temperature is 90°c what is the final temperature of the gold water system

89.88° C

We are given;

We are required to calculate the final temperature;

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

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

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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