Question

You cool a 100.0-g slug of red-hot iron (temperature ${\mathbf{745}}^{\mathbf{o}}\mathbf{}\mathit{C}$) by dropping it into an insulated cup of negligible mass containing 85.0 g of water at $\mathbf{20}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}\mathbf{}\mathit{C}$. Assuming no heat exchange with the surroundings, (a) what is the final temperature of the water and (b) what is the final mass of the iron and the remaining water?

Short answer

a) The final temperature of water is 100ºC

b) The final mass of iron and remaining water is 184.2 g

Step-by-step solution

(a) Determination of final temperature of the water.

Initially the iron is solid as the normal melting point of iron is more than 745°C

The total of the system is zero because the system is insulated,

${\mathit{Q}}_{\mathbf{s}\mathbf{y}\mathbf{s}\mathbf{t}\mathbf{e}\mathbf{m}\mathbf{}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}$

For water, specific heat is,

$\mathrm{c}=4190\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$

For water, Latent heat of vaporization is,

$L\_v=2256.10J/kg$

For solid iron, specific heat is,

$c\_iron=470J/kg\cdot K.$

The amount of heat released while iron slug cools to 100°C is,

${\mathbf{Q}}_{\mathbf{1}}\mathbf{=}{\mathbf{mc}}_{\mathbf{iron}}\mathbf{∆}\mathbf{T}$

$$\begin{gathered} =\left(0.1000\mathrm{kg}\right)\left(470\mathrm{J}/\mathrm{K}\right)\left(645\mathrm{K}\right) \\ =3.03\times {10}^4\mathrm{J} \end{gathered}$$

The amount of heat absorbed while the temperature of the water rises to 100°C is

localid="1664360063801" ${\mathbf{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{mc}\mathbf{∆}\mathbf{T}$

$$\begin{gathered} =\left(0.0850kg\right)\left(4190J/kg.k\right)\left(80.0K\right) \\ =2.85\times {10}^{4}J \end{gathered}$$

The amount of heat released is greater than the amount of heat absorbed. The amount of residual heat is,

$$\begin{gathered} Q_1-Q_2=3.03\times {10}^{4}J-2.85\times {10}^{4}J \\ =1.81\times {10}^{3}J \end{gathered}$$

This residual heat will convert some of the water mass to vapor. Thus, the maximum temperature achieved should be 100°C.

(b) Determination of the final mass of iron and remaining water.

Latent heat of vaporization expression is given as,

$\mathit{Q}\mathbf{=}\mathit{m}\mathit{L}$

$\mathit{m}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{L}}$

Substitute Q = $1.81\times {10}^{3}J$and $L\_v=2256.10J/kg$in the above equation,

$$\begin{gathered} m=\frac{1.81\times {10}^{3}J}{2256.10J/kg} \\ =8.01\times {10}^{-4}kg \\ =0.801g \end{gathered}$$

This is the mass of water lost after vaporization.

Therefore, Remaining mass of water is,

$85.0g-0.801g=84.2g$

Thus, the total mass of iron and remaining water is,

$100.0\hspace{0.33em}g+\hspace{0.33em}84.2\hspace{0.33em}g=184.2\hspace{0.33em}g$