Question

An electric heater used to boil small amounts of water consists of a 15-Ω coil that is immersed directly in the water. It operates from a 60-V socket. How much time is required for this heater to raise the temperature of 0.80 kg of water from 20° C to the normal boiling point?

Answer 1

In physics, the time required to boil water with an immersion heater can be calculated using energy, power, and Ohm's law formulas. By finding the energy needed and the power of the heater, one can determine the time it takes to heat the water to the boiling point.

Step-by-step explanation:

The subject of this question is Physics, specifically related to the concepts of electricity and thermodynamics. A 15-Ω coil that operates from a 60-V socket is used as an immersion heater to boil water.

To answer the question, we need to find out how much time is required for this heater to raise the temperature of 0.80 kg of water from 20°C to 100°C. The energy needed to heat the water can be calculated by using the specific heat capacity formula:

Energy (Q) = mass (m) × specific heat capacity (c) × temperature change (ΔT)

Substituting the specific heat capacity of water (4,184 J/kg°C), the mass of the water, and the temperature change, we can calculate Q (Energy).

Next, we use the electric power formula to find the power (P) of the heater,

Power (P) = voltage (V) × current (I)

where current (I) can be found using Ohm's Law:

I = V/R (where R is the resistance)

Once we have the power, we can find the time (t) it will take to heat the water by rearranging the formula for power:

Power (P) = Energy (Q) / time (t)

So, time (t) = Energy (Q) / Power (P)

Answer 2

t = 1120 seconds

Step-by-step explanation:

Law of Conservation of Energy: It states that energy can neither be created or destroyed but it can be transformed from one form to another

The heater converts electric energy to heat energy.

According to the law of conservation of Energy,

Heat supplied by the heater = heat gained by the water.

Heat supplied by the heater = (V²/R)t, Where V = voltage, R = Resistance, t = Time.

Heat gained by the water = cm(θ₂ -θ₁)

Where c = specific heat capacity of water, m = mass of water, θ₂= final temperature of water, θ₁= initial temperature of water.

∴ (V²/R)t = cm(θ₂ -θ₁)

Making t the subject of the equation,

t = cm(θ₂ -θ₁) ×R/V²

Where c= 4200 Jkg/K, m=0.80 kg, R = 15 Ω, V = 60 V, θ₁= 20°C, θ₂ = the normal boiling point of water = 100°C

∴ t = (4200×0.8)(100 -20)×15/60²

t = (3360×80×15)/3600

t = 4032000/3600 = 1120 seconds

∴ The time required by the heater to raise the temperature of water to normal boiling point is 1120 seconds.