60 g of hot water at 90°C mixed with 40

Berikut ini adalah pertanyaan dari hihusayaorangwaowwow pada mata pelajaran Fisika untuk jenjang Sekolah Menengah Pertama

60 g of hot water at 90°C mixed with 40 g of water at 25°C, if the specific heat capacity of water is 4.2 J/g.°С. What is the final temperature of the mixture?​

Jawaban dan Penjelasan

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Jawaban:

since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed.  Since the masses of the liquid at different temperatures are same, the answer is very easy and simple :  average of 20⁰C and 40⁰C.  that is:  30⁰C.

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final temperature of the mixture =

    = [ m1 * T1 + m2 * T2 ] / (m1 + m2)

    = [ 50 gms * 20⁰ C + 50 gms * 40⁰C ]  / (50+50)

    = 3,000 / 100 = 30⁰C

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another way using specific heats :

  let the final temperature be = T ⁰C

  Amount of heat given out by the hot water = m * s * (40⁰C - T)

          = 50 gms * s* (40 -T)

  Amount of heat taken in by the cold water = m * s * (T - 20⁰C)

      = 50 gms * s * (T - 20 )

  As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,

          50 * s * (40 -T) = 50 gm * s  * (T-20)

                40 - T = T - 20

            2 T = 60    => T = 30⁰C

Penjelasan:

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Last Update: Sun, 30 May 21