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24 March, 20:57

A 13.5 g sample of gold is heated to 125.0°C, then placed in a calorimeter containing 60.0 g of water. The final temperature of the water 20.00 o C. The specific heat of gold is 0.130 J/g o C. What was the initial temperature of the water?

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  1. 24 March, 21:11
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    The initial temperature was 19.27 °C.

    The guiding principle is the Law of Conservation of Energy: the sum of all the energy transfers must add up to zero.

    The formula for the heat q gained or lost by a substance is

    q = mCΔT

    where

    m = the mass of the substance.

    C = its specific heat capacity.

    ΔT = T_f - T_i = the change in temperature.

    In this problem, there are two heat transfers.

    Heat lost by gold + heat gained by water = 0

    m _1C_1ΔT_1 + m_2c_2ΔT_2 = 0

    m_1 = 13.5 g; C_1 = 0.130 J·°C^ (-1) g^ (-1); ΔT_1 = T_f - T_i = 20.00 °C - 125.0 °C = - 105.0 °C

    m_2 = 60.0 g; C_2 = 4.184 J·°C^ (-1) g^ (-1); ΔT_2 = ?

    q_1 = m_1C_1ΔT_1 = 13.5 g * 0.130 J·°C^ (-1) g^ (-1) * - 105.0 °C = - 184.3 J

    q_2 = m_2C_2ΔT_2 = 60.0 g * 4.184 J·°C^ (-1) g^ (-1) * ΔT_2

    = 251.0 ΔT_2 J·°C^ (-1)

    q_1 + q_2 = - 184.3 J + 251.0 ΔT_2 J·°C^ (-1) = 0

    251.0 ΔT_2 °C^ (-1) = 184.3

    ΔT_2 = 184.3/251.0 °C^ (-1) = 0.734°C

    ΔT_2 = T_f - T_i = 20.00 °C - T_i = 0.734 °C

    T_i = 20.00 °C - 0.734 °C = 19.27 °C
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