Question

An iron rod at 80°C of mass 0.8 kg is dropped into 1 kg of water at 20°C. What will be the final
temperature of rod and water in equilibrium?
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{PLEASE ANSWER IN STEP BY STEP EXPLANATION}
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Answer 1

Hey,

Lets consider mass of iron as m1=0.8kg

                         water mass m2=1kg

               temperature of iron t2=80*C

               temperature of water t2=20*C

According to law of conversation of energy, energy gained = energy lost.

Hence Q1 =Q2

   c of iron × m1 × t1-t = c of water × m2 ×t- t2

       448 * 0.8 * t1-t = 4186 * 1 * t-t2

       Now find answer by yourself I am tired of typing now its just numerical part.

Thank you