Before brewers had access to propane or natural gas, they would heat large, insulated vats of water by adding hot stones to the water. The stones are heated to 800°C before they are dropped into a particular well-insulated large vat holding 1,200 kg of water at 20°C. If we want to heat the water to 70°C, what mass of stones must be added the the vat? The specific heat of the stone is about 790 J/kg°C and the specific heat of water is 4,186 J/kg°C. Assume the mass of the water does not change and that the vat is a perfect insulator so no internal energy is lost to

$$\begin{gathered} Heat gained by water = mass of water *\hspace{0.17em}specific heat of water * change in temperature \\ = 1200 *\hspace{0.17em}4186 *\hspace{0.17em}(70-20) \\ = 251160000 Joules \\ Now, this amount of energy must be lost by stone \\ Thus , H_{stone} = - 251160000 = mass of stone * specific heat of stone * change in temperature of stone \\ -251160000 = - m *\hspace{0.17em}790 *\hspace{0.17em}(70 - 800) \\ m = \mathbf{435}\mathbf{.}\mathbf{5}\mathbf{ }\mathit{k}\mathit{g} \end{gathered}$$