Question

An aluminum baking sheet with a mass of 225 g absorbs 2.4 x 104 J from an oven. If its temperature was initially 25 C, what will its new temperature be?

Answer 1

Answer:

The value is $T_2 =416.9 \ K$

Explanation:

From the question we are told that

     The mass of the aluminum baking sheet is  $m = 225 \ g = 0.225 \ kg$

      The energy absorbed is $E = 2.4 *10^{4} \ J$

       The initial  temperature is  $T_1 = 25 ^oC = 25 + 273 = 298 \ K$

   

Generally the heat absorbed is mathematically represented as

         $Q = m * c_a * [T_2 - T_1]$

Here  $c_a$ is the specific heat capacity of  aluminum with value  $c_a = 897 \ J / kg \cdot K$

So

           $2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298]$

=>         $T_2 - 298 = 118.915$

=>          $T_2 =416.9 \ K$