Which mass of water will it heat up from ${20}^{\circ}\mathrm{C}$ to ${50}^{\circ}\mathrm{C}$ in the time of 10 minutes? (C of water = $4200\text{}\mathrm{J}\phantom{\rule{thickmathspace}{0ex}}\mathrm{k}\mathrm{g}/\mathrm{K}$)

icedagecs
2022-07-14
Answered

A heater made out of a wire with a diameter $R=0.2\text{mm}$, length $4\pi \text{m}$ and electrical resistivity of $0.5\times {10}^{-6}\text{}\mathrm{\Omega}\phantom{\rule{thickmathspace}{0ex}}\mathrm{m}$ is connected to a voltage source of $220\text{V}$, sinked in the water.

Which mass of water will it heat up from ${20}^{\circ}\mathrm{C}$ to ${50}^{\circ}\mathrm{C}$ in the time of 10 minutes? (C of water = $4200\text{}\mathrm{J}\phantom{\rule{thickmathspace}{0ex}}\mathrm{k}\mathrm{g}/\mathrm{K}$)

Which mass of water will it heat up from ${20}^{\circ}\mathrm{C}$ to ${50}^{\circ}\mathrm{C}$ in the time of 10 minutes? (C of water = $4200\text{}\mathrm{J}\phantom{\rule{thickmathspace}{0ex}}\mathrm{k}\mathrm{g}/\mathrm{K}$)

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escampetaq5

Answered 2022-07-15
Author has **12** answers

Find the resistance of the particular wire. Then calculate the power it uses. Assume this power is dissipated as heat. Find how much energy is converted to thermal energy by heat in 10 minutes. Use the equation $Q=mc\mathrm{\Delta}T$ to find the mass of the water.

ban1ka1u

Answered 2022-07-16
Author has **5** answers

You may consider this question from perspective of energy view: the electric energy is consumed by the resistor and convert this energy to the thermal energy (the source of heat that heat up the water). So from this point of view, if you can assume the 100% electric energy converting to thermal energy, and usually, this assumption is right for resistors since there is no other kind of energy that electric energy can convert to, since this is not a motor or a light bulb. Thus you will have the following equation:

$Heat={I}^{2}Rt=\frac{{U}^{2}}{R}t$

So this amount of heat will be absorbed by water and heat the water up, so

$Heat={c}_{water}\cdot m\cdot \mathrm{\Delta}T$

By solving these two equations, you can get the amount of water (${m}_{water}$) you need as

${m}_{water}=\frac{{U}^{2}t}{R\cdot c\cdot \mathrm{\Delta}T}$

where R can be easily calculated as $R=\frac{\rho L}{\pi {r}^{2}}$ for this cylindrical resistor.

So the key point to connect the electricity to thermal dynamics is the conservation of energy so that energy has to convert from one form (electric energy in your case) to another form (thermal energy or heat in your case), and little or no energy is converted into other form such as light.

$Heat={I}^{2}Rt=\frac{{U}^{2}}{R}t$

So this amount of heat will be absorbed by water and heat the water up, so

$Heat={c}_{water}\cdot m\cdot \mathrm{\Delta}T$

By solving these two equations, you can get the amount of water (${m}_{water}$) you need as

${m}_{water}=\frac{{U}^{2}t}{R\cdot c\cdot \mathrm{\Delta}T}$

where R can be easily calculated as $R=\frac{\rho L}{\pi {r}^{2}}$ for this cylindrical resistor.

So the key point to connect the electricity to thermal dynamics is the conservation of energy so that energy has to convert from one form (electric energy in your case) to another form (thermal energy or heat in your case), and little or no energy is converted into other form such as light.

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So, first of all, we must factorize the denominator:

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${V}_{P}-{V}_{O}={\int}_{P}^{O}E.dr$

The equation is in one dimension and P and O are points on the x axis.

The Electric field is oriented in the $+x$ direction ans has the formula $c\frac{1}{r}$ where c is a coefficient and r is the distance form the origin of the coordinate system.

The vector ${\overrightarrow{r}}_{P}-{\overrightarrow{r}}_{O}$ has the same direction as that of the electric field.

Now, Why do we put P below the integral and O above it? Can someone explain the equation for me?

The equation is in one dimension and P and O are points on the x axis.

The Electric field is oriented in the $+x$ direction ans has the formula $c\frac{1}{r}$ where c is a coefficient and r is the distance form the origin of the coordinate system.

The vector ${\overrightarrow{r}}_{P}-{\overrightarrow{r}}_{O}$ has the same direction as that of the electric field.

Now, Why do we put P below the integral and O above it? Can someone explain the equation for me?

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