Answer
240$J$$Kg^{-1}$$K^{-1}$
Work Step by Step
We are going to use the formula $Q$=$m$$c$$\Delta$$T$.
In the problem, we are given the change in temperature, $18^{\circ}$$C$. The formula $Q$=$m$$c$$\Delta$$T$ is used to determine the energy required to raise the temperature of a substance, given we know the specific heat of the substance.
In this problem, the energy, $Q$, is given and we are trying to find $c$. This can be done with basic algebraic manipulation of the formula. By dividing both sides by $m$ and $\Delta$$T$, we are left with $c$=$Q$/$m$$\Delta$T. The mass of the substance, $m$, can be found by dividing the weight by $g$, the acceleration due to gravity. For this problem, the mass was calculated to be approximately 2.898$Kg$.
Using the numbers we are given in the problem, $c$=$\frac{1.25x10^{4}}{(2.898)(18)}$.
We can calculate the specific heat of the substance to be approximately 240$J$$Kg^{-1}$$K^{-1}$.
