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16 January, 16:02

If you combine 290.0 mL 290.0 mL of water at 25.00 ∘ C 25.00 ∘C and 140.0 mL 140.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

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  1. 16 January, 16:30
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    The final temperature is 47.79 °C

    Explanation:

    Step 1: Data given

    Sample 1 has a volume of 290.0 mL

    Temperature of sample 1 = 25.00 °C

    Sample 2 has a volume of 140.00 mL

    Temperature of sample 2 = 95.00 °C

    Step 2: Calculate the final temperature

    Heat lost = heat gained

    Qlost = - Qgained

    Q = m*c * ΔT

    Q (sample1) = - Q (sample2)

    m (sample1) * c (sample1) * ΔT (sample1) = - m (sample2) * c (sample2) * ΔT (sample2)

    ⇒with m (sample1) = the mass of sample 1 = 290.0 mL * 1g/mL = 290 grams

    ⇒with c (sample 1) = the specific heat of water = c (sample 2)

    ⇒with ΔT (sample 1) = the change of temperature = T2 - T1 = T2 - 25.00 °C

    ⇒with m (sample2) = the mass of sample 2 = 140.0 mL * 1g/mL = 140 grams

    ⇒with c (sample2) = the specific heat of water = c (sample1)

    ⇒with ΔT (sample2) = T2 - T1 = T2 - 95.00°C

    m (sample1) * ΔT (sample1) = - m (sample2) * ΔT (sample2)

    290 * (T2-25.0) = - 140 * (T2 - 95.0)

    290 T2 - 7250 = - 140 T2 + 13300

    430 T2 = 20550

    T2 = 47.79 °C

    The final temperature is 47.79 °C
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