A certain brand of hot-dog cooker works by applying a potential difference of 120 Vacross opposite ends of a hot dog and allowing it to cook by means of the thermal energy produced. The current is 10.0 A, and the energy required to cook one hot dog is 60.0 kJ. If the rate at which energy is supplied is unchanged, how long will it take to cook three hot dogs simultaneously?

a) The potential difference,$V=120V$

b) The current flowing is$I=10.0A$

c) The energy required to cook one hot dog is$E_1=60.0\mathrm{kJ}$

The power or rate of energy dissipation, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

Here, we need to use the equation of power relating to energy and time. We have given the current and the potential, using these values we can calculate the power. From these two expressions, we can find the time required to cook three hot dogs simultaneously.

Formulae:

The electric power of the circuit,$P=IV$ …(i)

Here,P is the power,I is current,V is the potential difference.

The rate of energy consumed to produce heat,$P=\frac{E}{t}$ …(ii)

Here,P is the power,E is electric energy,t is the time.

Equating equations (i) and (ii),we can get the time taken by the cooker as follows:

$t=\frac{E}{I\times V}$ …(iii)

We have the energy required to cook one hot dog is,$E_1=60.0kJ$

So, the energy required to cook three hot dogs will be,

$$\begin{gathered} E=3\times E_1 \\ =3\times 60.0\mathrm{kJ} \\ =180\mathrm{kJ} \\ =1.8\times {10}^5\mathrm{J} \end{gathered}$$

Thus, substituting these values with the given data in equation (iii), we get the time taken as follows:

$$\begin{gathered} t=\frac{1.8\times {10}^5\mathrm{J}}{10.0\mathrm{A}\times 120\mathrm{V}} \\ =150\mathrm{s} \end{gathered}$$

Hence, the value of the time taken by the cooker to cook three hot dogs is 150s.