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7 April, 07:03

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the specific heat of the metal (0.900 j/g degrees Celsius), fund the final temperature of the block and the water

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  1. 7 April, 08:52
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    Final temperature is 34.2 °C

    Explanation:

    Given dа ta:

    mass of metal = 125 g

    temperature of metal = 93.2 °C

    mass of water v = 100 g

    temperature of water = 18.3 °C

    specific heat of meta is = 0.900 j/g. °C

    specific heat of water is = 4.186 j/g. °C

    final temperature of water and metal = ?

    Solution:

    Q = m. c. ΔT

    ΔT = T2-T1

    now we will put the values in equation

    Q1 = m. c. ΔT

    Q1 = 125 g. 0.900 j/g. °C. 93.2°C - T2

    Q1 = 112.5 (93.2°C - T2)

    Q1 = 10,485 - 112.5T2

    Q2 = m. c. ΔT

    Q2 = 100. 4.186. (T2 - 18.3)

    Q2 = 418.6. (T2 - 18.3)

    Q2 = 418.6T2 - 7660.38

    10,485 - 112.5T2 = 418.6T2 - 7660.38

    10,485 + 7660.38 = 418.6T2 + 112.5T2

    18145.38 = 531.1 T2

    T2 = 18145.38/531.1

    T2 = 34.2 °C
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