Question

What would be the final temperature of a mixture of 100g of water at 60°C temperature and 200g
of water at 40°C temperature? (ASI)​

Answer 1

Answer:

Specific Heat of water is defined as the amount of heat (added or subtracted) in order to change the temperature of 1 gram of water by 1 degrees Kelvin (or Celsius), under constant pressure.

Let's assume a constant value (c) for the Specific Heat of water, between freezing and boiling temperatures and at atmospheric pressure.

After mixing the water, the total of 300 grams will reach an equilibrium temperature T.

The water originally at 90 degrees temperature will "loose" an amount of heat equal to: Delta Q = c*200 [grams]*(90-T) [degrees].

This same amount of heat will be absorbed by the water originally at 30 degrees to raise its temperature to T. Delta Q = c*100*(T-30).

Now it is easy to solve this equality:

c*200*(90-T) = c*100*(T-30), giving T=70 degrees Celsius

Answer 2

Answer:

30⁰C

Explanation:

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