Physicist in the kitchen - run for your lives!

First, we work out the volume of the egg in $m^3$ using,

$V=\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2 cm)^3 = \frac{4}{3} \pi \times (0.02 m)^3$

$= \frac{4}{3} \pi \times 0.000008m = \frac {0.000032 \pi}{3}m$

$= \frac {3.2 \times 10^{-5} \times \pi}{3}m$

Now, we work out the mass in $kg$ using,

$m= \rho V=1000kg/m^3 \times \frac {3.2 \times 10^{-5} \times \pi}{3}m$

$= \frac {10^3 \times 3.2 \times 10^{-5} \times \pi}{3}kg$

$= \frac {3.2 \times 10^{-2} \times \pi}{3}kg$

Next, let's work out the amount of heat energy needed to raise the temperature from 50⁰C to 65⁰C using,

$E=mc \Delta T = \frac {3.2 \times 10^{-2} \times \pi}{3}kg \times 4200 \frac {J}{kg⋅K} \times (65⁰C-50⁰C)$

$= \frac {3.2 \times 10^{-2} \times \pi}{3}kg \times 4200 \frac {J}{kg⋅K} \times 15⁰K$

$= \frac {42 \times 10^2 \times 3.2 \times 10^{-2} \times \pi \times 15}{3}J$

$= 672 \pi J$

Finally, we work out the time in seconds to supply this amount of energy, using,

$P= \frac {E}{t} \Longrightarrow t= \frac {E}{P} = \frac {672 \pi J}{20 W}$

$= \frac {672 \pi J}{20 J s^{-1}} = \frac {672 \pi}{20} s = 33.6 \pi s = \frac {33.6 \pi}{60} min$

$= 0.56 \times \pi$ min $=1.759 min$ (approx.)

Hence, the required answer is $\fbox {1.759}$ .

We note that the time is just $t = \dfrac {E}{P}$ where $P$ is the given power and $E$ is the heat energy, given by $mc \cdot \Delta T$ , where $c$ is the given specific heat and $\Delta T = 15 \, \text{C}^\circ$ . Thus: $t = \dfrac {mc \cdot \Delta T}{P}.$ But note that we need to find $m$ using the volume density and the volume as follows: $m = \rho V = \dfrac {4}{3} \pi r^3 \rho,$ since the egg is a sphere. Thus: $t = \dfrac {4}{3} \cdot \dfrac {\pi r^3 \rho c \cdot \Delta T}{P}.$ We have the necessary number to substitute into this. So we proceed.

$\implies r = 0.02 \, \text{m}$

$\implies \rho = 1000 \, \text{kg}/\text{m}^3$

$\implies \Delta T = 15 \, \text{C}^\circ$

$\implies P = \text {20 W}$

Plugging in, we get $t = 105.56 \, \text{s}$ . Converting to minutes, we get $\boxed {1.76 \, \text{minutes}}$ .

We need to calculate the mass of the egg, $M=\rho\frac{4}{3}\pi r^3$ and then the amount of energy required to raise the temperature by 15 degrees is $E = 15Mc$ . Since we want the time, we divide by the power and convert from seconds to minutes: $t = \frac{15Mc}{(20)(60)} = \frac{15\rho\frac{4}{3}\pi r^3c}{(20)(60)} = \frac{15(1000)( \frac{4}{3}\pi (.02)^3 (4200)}{1200} = 1.759$ .

We know that, the power, $P$ is how many energy store in a seconds, $\frac{E}{t}$ . So, to get the times in minutes , $t$ , You have to find the energy, energy in here mean the calor/heat enery which $E = mc\Delta T$ , $m$ mass you can get using $\rho = \frac{m}{V}$ . Volume in here can be modeled as sphere of radius $2 cm$ . Then,

$V = \frac{4}{3}\pi r^{3}$ , you will get the mass. Next, find the energy, $E$ , and $\Delta T$ is the temperature change in $^{\circ} C$ . Finally, you will the get time in seconds, using the relation of (\E) and $P$ . Converse it in to minutes .

There are lots of assumptions made in this question. In reality, there is a temperature gradient between the outer part of the egg and the (yolk) in the middle, which means that the inner part of the egg is always cooler than the outer part.

This gradient diminishes with time, which makes it a time dependent problem, solvable using a differential equation.

However, (intended) simplifications make the problem solvable.

By considering 1. a constant input of heat to the egg, and 2. the egg temperature is uniform, then we only have a single temperature to contend with, which varies linearly with time.

mass of egg: $m$ = $\rho(4/3)\pi r^{3}$ =1 $g/cm^{3}$ * $(4/3)\pi (2^{3})$ = $32\pi/3 g$ = $32\pi/3000 kg$

Heat capacity $C$ = $m (kg) *4200 J-kg/°K$ (given) = $224 \pi/ 5 J/°K$

Power supplied, $P$ = $20W$

Temp difference, $\Delta T$ = $(65-50)$ = $15 °K$

Time required = $\Delta T * C/P °K * J/°K * s/J$ = $15*(224\pi/5)/20 s$ = $105.6 s$ about 1 minute 45 seconds or $1.75 minutes$

$\begin{cases} \begin{aligned} \Delta T &= 65 - 50 = 15\text{K}\\ m_{egg} &= V_{egg} \cdot \rho_{egg} = \frac{4}{3} \pi (0.02)^3 \cdot 1000 \approx 0.0335\text{kg}\\ c &= 4200 \frac{J}{kg \cdot K} \end{aligned} \end{cases}$

Filling it in:

$Q = c \cdot m \cdot \Delta T = 4200 \cdot 0.0335 \cdot 15 = 2110.5\text{J}$

Finally, $\frac{2110.5 \text{J}}{20 \text{W}} \approx 105.5\text{s} \approx \boxed{1.76}\text{ minutes}.$

The enthalpy change of the egg is described by:

$E=mc\Delta\theta$

Differentiating with respect to t, and remembering that $\frac{dE}{dt}=P$ :

$P=\frac{d(mc\theta)}{dt}$

Because mass and secific heat capacity remain constant as t changes, they can be brought outside of the differential, meaning that:

$P=mc\frac{d\theta}{dt}$

Rearranging to find $d\theta$ :

$\frac{P}{mc}dt=d\theta$

Integrating both sides:

$\int\frac{P}{mc}dt=\int{d\theta}$

The lower bound for the left side is when $t=0$ , and the upper bound is when $t=t_1$ , where $t_1$ is the value to be determined. The lower bound for the right side is when $\theta=50$ , and the upper bound is when $\theta=65$ . This means that the resulting integral is:

$\int_{0}^{t_1}\frac{P}{mc}dt=\int_{50}^{65}{d\theta}$

Integrating both sides:

$[\frac{P}{mc}t]_{0}^{t_1}=[\theta]_{50}^{65}$ (1)

We'll leave this equation to one side for now. Mass can be described by density multiplied by volume, so:

$m={\rho}V$

$\rho$ is given, however:

$V=\frac{4}{3}{\pi}r^3$

This means that:

$m=\frac{4}{3}{\pi}{\rho}r^3$

Plugging that into equation (1) means that equation (1) is:

$[\frac{P}{\frac{4}{3}{\pi}c{\rho}r^3}t]_{0}^{t_1}=[\theta]_{50}^{65}$

$\frac{P}{\frac{4}{3}{\pi}c{\rho}r^3}$ is constant as t varies, so by plugging in $P=20$ , $c=4200$ , ${\rho}=1000$ and $r=0.02$ :

$\frac{P}{\frac{4}{3}{\pi}c{\rho}r^3}=\frac{25}{56\pi}$

Meaning that equation (1) is:

$[\frac{25}{56\pi}t]_{0}^{t_1}=[\theta]_{50}^{65}$

Evaluating bounds:

$\frac{25}{56\pi}t_1=65-50=15$

This means that:

$t_1=\frac{15\times56\pi}{25}$ in seconds, or:

$t_1=\frac{15\times56\pi}{25\times60}$ in minutes. Therefore:

$t_1=\frac{14}{25}\pi$ minutes, or, to 3 significant figures: $t_1=1.76$ minutes.

Let $Q$ be the heat needed for the egg to turn from soft to hard-boiled, $m$ the egg's mass, $d$ the egg's density, $V$ the egg's volume, $r$ the egg's radius, $ΔT$ the temperature difference between the two states of the egg and $Δt$ the desired time interval.We know that $Q=mcΔT$ with $Q=PΔt$ and $m=dV=d\frac{4}{3}πr^{3}$ .Substituting these into the first equation and solving for $Δt$ we get that $Δt=\frac{4πdcΔTr^{3}}{3P}=105.56 sec=1.759 min$

This problem can be solved by simple dimensional analysis: $\dfrac{1 s}{20 J} \times \dfrac{1 min}{60 s} \times \dfrac{4200 J}{kg \times K} \times 15K \times \dfrac{1000 kg}{m^3} \times \dfrac{4}{3}\pi (0.02m)^3 = 1.759 min$ Notes: 1 Watt = 1 J/s, the increments between the celsius and kelvin scales are equal 65 ºC - 50 ºC = 15 K.

We have the radius of the egg, $r=0.02$ meters. Since the egg can be modeled as a sphere, we can find the volume of the egg using the formula $V=\frac{4}{3}\pi\times r^3$ .

We also have the density of the egg, $\rho=1000$ $kg/m^3$ .

Multiplying the volume & the density of the egg gives us the mass, $m=\rho V$ .

As a soft boiled egg becomes hard-boiled, the temperature changes from $50^\circ C$ to $65^\circ C$ . In other words, the change of temperature that takes place is equal to $\Delta\theta=(65-50)$ or $15$ kelvins.

So the amount of heat needed for a soft-boiled egg to become hard boiled is equal to

$mc\Delta\theta=Q$ .

Dividing $Q$ by $P$ gives us the time $t$ .

In other words, $t=\frac{Q}{P}$ .

Wait! notice that $t$ is now in seconds and we want $t$ to be in minutes . So we need to divide our answer by $60$ to get our desired result.

$t_{minutes}=\frac{Q}{P\times60}=\frac{\rho V c \Delta\theta}{P\times 60}$ .

Now simply plug in all the values (use a calculator for this) and get $t=1.759$ minutes.

PS: A quick aside. I was wondering: is $c$ really the heat capacity of the egg? Looking at its unit tells me that it's the specific heat capacity of the composition of the egg and we were always taught at school to make this distinction.