Question

A 20 gram iron block at an initial temperature of 20oC is placed in 20 grams of water at an initial temperature of 80oC. Assuming no heat is lost to the surroundings, what would be the final temperature of the system once you reach thermal equilibrium?

CSiron=0.449JgoC

CSwater=4.18JgoC

74.2o C

80.0o C

25.8o C

50.0o C

112o C

Answer 1

Let the temperature at thermal equilibrium = X⁰C

When iron (colder) comes in contact with water (hot), heat flows from water to iron. Thus, at equilibrium, iron is at higher temperature than the initial and water is at lower temperature than its initial.

Thus, temperature of iron at equilibrium = X- 20⁰C

And, temperature of water at equilibrium = 80⁰C – X

Now, using q = m x c x dT                  

Where, q= heat gained/ lost

m= mass in gram,

c= specific heat

Also, heat gained by iron = heat lost by water

So,

                m1 c1 dT (iron) = m2 c2 dT (water)

                or, 20 g x 0.449 J/ g⁰C x (X- 20⁰C) = 20 gm x 4.18 J/ g⁰C x (80⁰C - X) =

                or, 8.98 J/ ⁰C x (X- 20⁰C) = 83.6 J/ ⁰C x (80⁰C - X)

                Or, [8.98 J/ ⁰C x (X- 20⁰C)] / 83.6 J/ ⁰C = (80⁰C - X)

                Or, 0.1074162679 x (X- 20⁰C) = (80⁰C - X)

                Or, (0.1074162679 X) – (2.14832535885⁰C) = 80⁰C – X

Or, 1.1074162679 X = 80⁰C + 2.14832535885⁰C = 82.14832535885⁰C

Or, X = 74.18⁰C

Hence, the temperature at thermal equilibrium = 74.18⁰C

Correct option = a