Question

An object of mass 3 kg is initially at a temperature of 400K. If the specific heat of this object is 478 J/kg * K, and 13 kJ of heat are added to the object, what is its final temperature?

Answer 1

Given,

The mass of an object is m = 3 kg

Initial temperature, T1 = 400 K

The specific heat of this object is c = 478 J/kg.K

The heat is Q = 13 kJ

Let T2 be the final temperature.

The formula is used to calculate the final temperature.

$\begin{gathered} Q=mc\Delta T \\ 13\text{ kJ=3 }kg\text{ }\times478\text{ J/}kg\cdot K\text{ }\times(T_2-400\text{)} \\ (T_2-400)=\frac{13\times10^3}{3\times478}_{} \\ T_2-400=9.065 \end{gathered}$

Thus, the value of the final temperature is

$\begin{gathered} T_2=400+9.065 \\ T_2=409.065\text{ K} \end{gathered}$

Therefore, the given answer (a) is correct.