Question

A chunk of steel with a mass of 2.3 kg absorbs net thermal energy of 2.5×105
J and rise its temperature by 350°C. What is the specific heat of the steel if the steel was at 27°C.?

Answer 1

Answer:  Specific Heat Capacity C = 336.53 J Kg⁻¹K⁻¹

Explanation:

Using Relationship between Heat Transfer (Q) and temperature change (ΔT)

Q= mcΔT

Here m- mass of the material

        c- specific heat of the material

    It can be written as

                                  c = Q/ m ΔT

Initial temperature Ti = 27°C =300K

Final Temperature Tf = 350°C = 623K

change ΔT = Tf - Ti =323K

              Put all the given values in equation (1)

                      c = 2.5 * 10⁵ J/ (2.3 kg * 323K)

                       c = 336.53 J Kg⁻¹ K⁻¹