# A 35 g sample of Fe is dropped into 80 g of water at 20 degrees Celsius. What is the final...

## Question:

A 35 g sample of Fe is dropped into 80 g of water at 20 degrees Celsius. What is the final (equilibrium) temperature of the iron/water?

## Conservation of Energy:

• Assuming a hot object at initial temperature ({eq}\textrm{T}_{1} {/eq}) is brought in contact with another object at a temperature {eq}\textrm{T}_{2} {/eq} such that {eq}\textrm{T}_{2} {/eq}<{eq}\textrm{T}_{1} {/eq}.
• Both bodies will attain a common temperature after some time as heat lost by the first object will be equal to the heat gained by another object. Energy is always conserved during any process.

## Answer and Explanation:

Given,

• Amount of water = m = 80 g
• Initial temperature of water = {eq}\textrm{T}_{1} {/eq} = {eq}20^{\circ}C {/eq}
• Specific heat capacity of water = c = 4.184 {eq}\textrm{J/g}^{\circ}C {/eq}
• Specific heat capacity of Fe = c' = 0.450 {eq}\textrm{J/g}^{\circ}C {/eq}
• Amount of Fe = m' = 35 g
• Assuming initial temperature of iron ({eq}\textrm{T}_{2} {/eq}) = {eq}60^{\circ}C {/eq}
• Suppose the final temperature of the solution be {eq}\textrm{T}_{f} {/eq}

Heat lost by Fe = Heat gained by water

{eq}\begin{align*} \textrm{mc}\left ( 60-\textrm{T}_{f} \right )^{\circ}C&=m'c'\left ( T_{f}-20 \right )^{\circ}C \\ 35 \textrm{ g}\times 0.450 \textrm{ J/g}^{\circ}C\times \left ( 60-\textrm{T}_{f} \right )^{\circ}C&=80\textrm{ g}\times 4.184 \textrm{ J/g}^{\circ}C\times \left ( T_{f}-20 \right )^{\circ}C \\ \textrm{T}_{f}&=21.79^{\circ}C \end{align*} \\ \boxed{\textrm{Final temperature of solution} = 21.79^{\circ}C} {/eq}