Question

What will be the range of the temperature of the mixture upon mixing

water at 400C with water at 60 0C ?




Please right answer dena .​

Answer 1

The mixture will attain Thermal Equilibrium after some point of time. Let that equilibrium Tem,perature be T.

Heat lost by water at 40degree C (313K) to attain Temperature T, Q1 = M.Cp.dT = 0.075x4180x(313-T)

Heat gained by water at 20degree C (293K) to attain Temperature T, Q2 = M.Cp.dT = 0.050x4180x(T-293)

Since the system attains euilibrium, heat lost must be equal to heat gained.

Thus equating the above two equations i.e Q1 = Q2, we get T = 305K = 32 degree C

Answer 2

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

hope it helps you ✌✅✅✅