Question

Galen, the great physician of antiquity, suggested scaling temperature from a reference point defined by mixing equal masses of ice and boiling water in an insulated container. Imagine that this is done with the ice at $\mathbf{0}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}\mathit{C}$ andthe water at ${\mathbf{100}}^{\mathbf{0}}\mathit{C}$. Assume that the heat capacity of the container is negligible, and that it takes 333.4 J of heat to melt 1.000 g ice at$\mathbf{0}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}\mathit{C}$ to water at $\mathbf{0}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}\mathit{C}$. Compute Galen’s reference temperature in degrees Celsius.

Short answer

The final temperature is the Galen’s reference temperature which is $10.{12}^{0}C$.

Step-by-step solution

Heat capacity

Another term used to denote heat capacity isthermal capacity. It belongs to the category of intensive property. ItsSI unit isJ/K.

Computing Galen’s reference temperature in degree cesius

Let the final temperature reached by the system be ${\text{T}}_{\text{f}}$. The heat energy used for melting ice at $0.{00}^{0}C$to water at $0.{00}^{0}C$and warming the water from $0.{00}^{0}C$is equal to the heat energy given out by the cooling of water from $100.{00}^{0}C$. The relation can be given as:

$\text{- mass×specific heat of water∆T = mass×}{\left(\text{heat of fusion}\right)}_{\text{ice}}\text{+ mass×specific heat of water}$

$\text{- specific heat of water×∆T =}{\left(\text{heat of fusion}\right)}_{\text{ice}}\text{+ specific heat of water×∆T}$

Substituting the values in the equation:

$$\begin{gathered} \text{- specific heat of water×∆T =}{\left(\text{heat of fusion}\right)}_{\text{ice}}\text{+ specific heat of water×∆T} \\ \text{-}\left({\text{4.18J}}^0{\text{C}}^{\text{- 1}}{\text{g}}^{\text{- 1}}\right)\left({\text{T}}_{\text{f}}{\text{- 100}}^0\text{C}\right)\text{=}\frac{\text{333.4J}}{\text{g}}{\text{+ 4.18J}}^0{\text{C}}^{\text{- 1}}{\text{g}}^{\text{- 1}}\left({\text{T}}_{\text{f}}{\text{- 0.0}}^0\text{C}\right) \\ {\text{- 4.187T}}_{\text{f}}{\text{+ 418 = 333.4 + 4.18T}}_{\text{f}} \end{gathered}$$

On further solving,

$$\begin{gathered} {\text{T}}_{\text{f}}\text{=}\frac{{\text{84.6}}^0\text{C}}{\text{8.36}} \\ {\text{= 10.12}}^0\text{C} \end{gathered}$$

The final temperature is the Galen’s reference temperature which is $10.{12}^{0}C$