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

The value is
`T_2 =416.9 \ K`

Step-by-step 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`