Question

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piece has a mass of 60 kg and its specific heat is 0.1027 kcal/(kg · °C). He uses 810 kg of oil at 35°C. The specific heat of oil is 0.7167 kcal/(kg · °C). Once the metal is immersed in the oil, the temperature reaches an equilibrium value of 39°C. How hot was the forged metal piece just before he plunged it into the oil?

Answer 1

Answer:

The temperature of the metal is  $T_m = 376.8 ^o C$

Explanation:

From the question we are told that

     The mass of the metal is  $M = 60 \ kg$

     The specific heat of the metal is  $c_p = 0.1027 kcal/(kg \cdot ^oC)$

       The mass of the oil is $M_o = 810 \ kg$

       The temperature of the oil is  $T_o = 35^oC$

       The specific heat of oil is  $c_o = 0.7167 kcal/(kg \cdot ^oC )$

       The equilibrium temperature is $T_e = 39 ^oC$

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               $Q = - Mc_p \Delta T$

=>            $Q = -Mc_p (T_m - T_c)$

Where $T_ m$  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           $Q = M_o c_o \Delta T$

=>        $Q = M_o c_o (T_c - T_o)$

So  

         $Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)$

substituting values

          $- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)$

=>       $T_m = 376.8 ^o C$