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17 July, 09:33

A 10.0-gram sample of liquid water at 23.0°C absorbs 209 joules of heat. What is the final temperature of the liquid water sample?

a) 5.0°C b) 28.0°C c) 18.0°C d) 50.0°C

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  1. 17 July, 11:25
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    The final temperature of the water is 28.0 °C

    Explanation:

    Step 1: Data given

    Mass of liquid water = 10.0 grams

    Temperature = 23.0 °C

    Heat absorbed = 209 Joules

    Since heat was absorbed by the water, you must have a positive value for

    Δ T

    Step 2: Calculate final temperature

    q = m*c * ΔT

    ⇒ with m = the mass of the water = 10.0 grams

    ⇒ with c = the specific heat of water = 4.184 J/g°C

    ⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 23.0 °C

    ⇒ with q = the heat absorbed = 209 Joule

    209 = 10.0 * 4.184 * ΔT

    ΔT = 5

    ΔT = 5 = T2 - 23

    T2 = 28 °C

    The final temperature of the water is 28.0 °C
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