Question

An open container holds 0.550kg of ice at $\mathbf{-}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$. The mass of the container can be ignored. Heat is supplied to the container at the constant rate 800.0 J/min of for 500.0 min. (a) After how many minutes does the ice start to melt? (b) After how many minutes, from the time when the heating is first started, does the temperature begin to rise above $\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$? (c) Plot a curve showing the temperature as a function of the elapsed time.

Short answer

(a) t = 21.7 min

(b) t = 251.7 min

(c)

Step-by-step solution

Calculate the time using the heat flow equation

The heat flow equation is $\mathit{Q}\mathbf{=}\mathit{m}\mathit{c}\mathbf{∆}\mathit{T}$.

Given that ${\mathrm{m}}_{\mathrm{ice}}=0.550\mathrm{kg},{\mathrm{T}}_{\mathrm{ice}}=-15°\mathrm{C}\mathrm{and}\frac{∆\mathrm{Q}}{∆\mathrm{t}}=800\mathrm{J}/\min$.

$$\begin{gathered} \mathrm{Q}=\mathrm{mc\delta T} \\ \mathrm{Q}=0.550\times 2110\times (0+15) \\ \mathrm{Q}=1.73\times {10}^4\mathrm{J} \end{gathered}$$

By solving the rate localid="1665117914592" $\frac{∆Q}{∆t}$for time t

$$\begin{gathered} t=\frac{Q}{\frac{\Delta Q}{\Delta t}} \\ t=\frac{1.73\times {10}^4}{800} \\ t=21.7min \end{gathered}$$

So, the ice started to melt after 21.7 min.

Calculate the time for vaporisation

The heat involved in the vaporization process is

$$\begin{gathered} Q=m{L}_f \\ Q=0.55\times 3.34\times {10}^5 \\ Q=1.84\times {10}^{5}J \end{gathered}$$

By solving the time for the rate

$$\begin{gathered} t_2=\frac{Q}{\frac{\Delta Q}{\Delta t}} \\ {t}_2=\frac{1.84\times {10}^5}{800} \\ {t}_2=230min \end{gathered}$$

So, by summing the minutes at which ice starts melting and the time at which solid converts to liquid, we will get the total time

$$\begin{gathered} t_1+t_2=21.7+230 \\ {t}_1+t_2=251.7min \end{gathered}$$

Plot the graph

The graph of the process will be