Question

A 0.3 kg metal bar initially at 1200 k is removed from an oven and quenched by immersing it in a closed tank containing 9 kg of water initially at 300 k. Each substance can be modelled as incompressible. An appropriate constant specific heat value for the water is 4.2 kj/kg k and for the metal is 0.42 kj/kg k. Heat transfer from the tank contents can be neglected. Determine (a) the final equilibrium temperature (k) of the metal bar and the water

Answer 1

Answer:

302.99 K

Explanation:

Given = a) mass of metal bar = 0.3 kg

            b) specific heat of metal(s1) = 0.42 kl/kg k

            c) initial temperature of metal bar T1 = 1200 K

            d) specific heat of water(s2) = 4.2kj/kgk

            e) mass of water = 9 kg

            f) initial temperature of  water = 300K

FORMULA : ΔQ = MSΔT

HERE m = mass

         s = specific heat

        ΔT = change in temperature

       Δ Q = change in heat i.e heat gained or loss

we know that energy is neither created  not destroyed only change from 1 form to another

∴ heat lost by metal when dipped in water will be gained by water to increase temperature , so at equilibrium ΔQ will be same

⇒ ms1ΔT1 = ms2ΔT2

PUTTING THE VALUE ;

0.3 ×0.42×(1200 - Tfinal ) = 9×4.2×(Tfinal - 300)

⇒ 151.2 - 0.126x =  37.8x - 11340

37.926x = 11491.2

⇒ x = 11491.2/37.926 = 302.99 K

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