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15 September, 10:52

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

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  1. 15 September, 11:16
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    89.88° C

    Explanation:

    We are given;

    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 Step 1: Calculate the quantity of heat absorbed by the Gold cylinder

    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

    Step 2: Calculate the quantity of heat released by water

    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

    Step 3: Calculate the final temperature, X°C

    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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