# Heat Transfer

This wiki is the subject of a wiki collaboration party set to be held on Saturday, April 9th at 8:30pm IST (8:00am PST). Please add examples to the appropriate headings or under the examples heading.

Contribute wherever you would like! Here are the author breakdowns:

- Types of heat transfer: Sravanth/Julien
- Phases of matter: Ashish
- Steady-state heat transfer: Ameya
- Specific heat: Ashish
- Calorimetry
- Applications (green house effect, thermos): Sravanth

#### Contents

## Examples for Wiki Collaboration

How much heat energy is gained when $5\text{ kg}$ of water at $10\ ^\circ\text{C}$ is brought to its boiling point?

(Heat capacity of water) = $4200\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}$.

We have

$\begin{aligned} \text{Rise in temperature} ({\theta}_R) & = (100 - 10)\ ^\circ\text{C}\\ & = 90\ ^\circ\text{C}\\ \\ \text{Rise in temperature} & = m × c × {\theta}_R\\ & = 5 × 4200 × 90\\ & = 1890000 \text{ J}. \ _\square \end{aligned}$

A waterfall is $840\text{ m}$ high. If the initial temperature of water at the top of the waterfall is $15\ ^\circ\text{C}$, what is its temperature when it reaches the bottom?

Assume that all the energy of falling water changes into heat energy. Take $g = 10\text{ m/s}^2$ and specific heat capacity of water as $4200\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}$.

We have

$\begin{aligned} (\text{K.E. of falling water at the bottom}) & = (\text{P.E. of water at the top})\\ & = m × g × h\\ & = (\text{Heat energy produced by water})\\ \\ m × g × h & = m × c × {\theta}_R\\ \\ \Rightarrow {\theta}_R & = \dfrac{g × h}{c}\\ & = \dfrac{10 × 840}{4200}\\ & = 2\ ^\circ\text{C}\\ \\ (\text{Temperature of water at the bottom}) & = (15 + 2)\ ^\circ\text{C}\\ & = 17\ ^\circ\text{C}. \ _\square \end{aligned}$

## Types of Heat Transfer

Transfer of heat occurs in three different ways. As discussed earlier, a physical object can be transferred via three modes, but even energy can be transferred in similar modes.

**Conduction**

Conduction is the transfer of heat which happens through the particles in the medium as a result of a series of collisions, without the actual movement of particles, i.e. the particles do not act as the carriers of energy but they

passthe energy with the help of vibrations.Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference.

**Convection**

**Radiation**

## Phases of Matter

Main article: Phase changes

How much heat energy is required to melt $5 × {10}^6\text{ kg}$ of iron?

$\big($Specific latent heat (SLH) of iron$\big)$ = $2.7 × {10}^5\text{ J/kg}.$

We have

$\begin{aligned} (\text{Heat energy required to melt iron}) & = m × L\\ & = 5 × {10}^6 × 2.7 × {10}^5\\ & = 13.5 × {10}^{11}\\ & = 1.35 × {10}^{12}. \ _\square \end{aligned}$

$70\text{ g}$ of ice at $-30\ ^\circ\text{C}$ is heated by a burner such that it forms an equal amount of water at $40\ ^\circ\text{C}$. Find the amount of heat energy that should be supplied for this to happen.

- (SHC of ice) = $2.1\text{ J}/g\ ^\circ\text{C},$ (SLH of ice) = $336\text{ J}/g,$ and (SHC of water) = $4.2\text{ J}/g\ ^\circ\text{C}$.

We have

$\begin{aligned} (\text{Heat energy required to raise the temperature of ice to 0°C}) & = m × c × {\theta}_R\\ & = 70 × 2.1 × 30\\ & = 4410\text{ J}\\ \\ (\text{Heat energy required to melt the ice}) & = 70 × 336\\ & = 23520\text{ J}\\ \\ (\text{Heat energy required to raise the temp of ice at 0°C to water at 40°C}) & = m × c × {\theta}_R\\ & = 11760\text{ J}\\ \\ \Rightarrow (\text{Total heat energy required}) & = (4410 + 23520 + 11760)\text{ J}\\ & = 39690 \text{ J}.\ _\square \end{aligned}$

$300\text{ g}$ of ice at $-25\ ^\circ\text{C}$ is contained in a vessel of mass $10\text{ g}$ and specific heat capacity $0.4\text{ J}/g\ ^\circ\text{C}$. How much water at $50\ ^\circ\text{C}$ should be added to the vessel such that the final temperature of the combination is $15\ ^\circ\text{C}?$

- Assume that the vessel is at the same initial temperature as that of the ice contained in it.
- (SLH of ice) = $336\text{ J}/g,$ (SHC of ice) = $2.1\text{ J}/g\ ^\circ\text{C},$ and (SHC of water) = $4.2\text{ J}/g\ ^\circ\text{C}.$

Let amount of water required be $x.$ Then

$\begin{aligned} (\text{Amount of heat lost by water}) & = m × c × {\theta}_F\\ & = x × 4.2 × 35\\ & = 147x\\ \\ (\text{Heat energy required to raise the temperature of ice to 0°C}) & = m × c × {\theta}_R\\ & = 300 × 2.1 × 25\\ & = 15750\text{ J}\\ \\ (\text{Heat energy required to melt the ice}) & = 300 × 336\\ & = 10080\text{ J}\\\\ (\text{Heat energy required to raise temp. of ice at 0°C to water at 15°C}) & = m × c × {\theta}_R\\ & = 18900\text{ J}\\ \\ (\text{Heat energy required to raise the temp. of vessel}) & = m × c × {\theta}_R\\ & = 10 × 0.4 × (25 + 15)\\ & = 160\text{ J}. \end{aligned}$

According to the law of conservation of energy, heat gained is equal to heat lost. Thus,

$\begin{aligned} 147x & = (15750 + 10080 + 18900 + 160)\text{ J}\\ & = 44890\text{ J}\\ x & = 305.37415\text{ J}.\ _\square \end{aligned}$

A solid initially at $20\ ^\circ\text{C}$ is heated. The graph shows variation in temperature with the amount of heat energy supplied to it. Now, if the specific heat capacity of the solid is $2\text{ J}{g}^{-1}\ ^\circ\text{C}^{-1},$ calculate

(i) the mass of the solid and

(ii) the specific latent heat of fusion of the solid.

Submit your answer as the sum of (i) (in grams) and (ii) $($in $\text{ J/g}).$

$500\text{ g}$ of a metal of specific heat capacity $0.25 \text{ Jg}^{-1}\ ^\circ\text{C}^{-1}$ at $125\ ^\circ\text{C}$ is added to an equal amount of ice at $0\ ^\circ\text{C}$. Did all the ice melt?

If so, enter 555.

If not, enter the quantity of unmelted ice in grams (up to 2 decimal places).

**Note:** Specific latent heat of ice = $336\text{ Jg}^{-1}.$

## Steady State Heat Transfer

## Specific Heat

Main article: Specific heat

$\text{Q = Cm}\Delta\text T$

How much heat energy is gained when $5\text{ kg}$ of water at $10\text{ }^\circ\text{C}$ is brought to its boiling point?

Specific heat capacity of water is equal to $4200 \text{ J kg}^{-1}{\text{ }^\circ\text{C}}^{-1}.$

We have

$\begin{aligned} \text{Rise in temperature}({\theta}_R) & = (100 - 10)\text{ }^\circ\text{C}\\ & = 90\text{ }^\circ\text{C}\\ \\ \Rightarrow (\text{Rise in temperature}) & = m × c × {\theta}_R\\ & = 5 × 4200 × 90\\ & = 1890000\text{ J}. \ _\square \end{aligned}$

The heat capacity of a solid of mass $200\text{ g}$ is $500\text{ J}/ \, ^\circ\text{C}$. Calculate the specific heat capacity of the solid.

We have

$\begin{aligned} (\text{Heat capacity}) & = \text{m × c}\\ 500 & = 200 × c\\ \Rightarrow c & = \dfrac{500}{200}\\ & = 2.5 \text{ Jg}^{-1}{\text{ }^\circ\text{C}}^{-1}. \ _\square \end{aligned}$

Water at $80\text{ }^\circ\text{C}$ is poured in a bucket which contains $5\text{ kg}$ of crushed ice, such that all the ice melts and the final temperature recorded is $0\text{ }^\circ\text{C}$. Calculate the amount of hot water added to ice.

Take specific heat capacity of water is $4200\text{ J kg}^{-1}{\text{ }^\circ\text{C}}^{-1}.$ Also, take specific latent heat of fusion of ice is $336000\text{ J kg}^{-1}$.

Let the mass of hot water be $x\text{ kg}.$ Then

$\begin{aligned} (\text{Heat given out by hot water}) & = m × c × {\theta}_F\\ & = x × 4200 × 80\\ & = 336000 x\text{ J} \\ \\ (\text{Heat absorbed by ice to form water at 0°C}) & = m × L\\ & = 5 × 336000\\ & = 1680000\text{ J}. \end{aligned}$

By principle of calorimetry,

$\begin{aligned} (\text{Heat given out by hot water}) & = (\text{Heat absorbed by cold water})\\ 336000x &= 1680000\\ \\ \Rightarrow x & = \dfrac{1680000}{336000} =5 \text{ (kg)}.\ _\square \end{aligned}$

$100\text{ g}$ of solid $A$ of specific heat capacity $50\text{ J/g}\, ^\circ\text{C}$ at $90\ ^\circ\text{C}$ is placed in a superconducting vessel of mass $10\text { g}$ of specific heat capacity $1\text { J/g}\, ^\circ\text{C}$. Then some amount of water at $40\ ^\circ\text{C}$ is added to the vessel to cool the mixture to $55\ ^\circ\text{C}$.

Now, the above mixture after cooling down obtains a specific heat capacity of $25\text{ J/g}\, ^\circ\text{C}$. Then, to the above mixture is added a solid $B$ of specific heat capacity $20 \text{ J/g}\, ^\circ\text{C}$ at $10\ ^\circ\text{C}$ such that the final temperature of the overall combination becomes $30\ ^\circ\text{C}$.

How much amount of solid $B$ is added (up to 2 decimal places)?

## Calorimetry

Main article: Calorimetry

$100\text{ g}$ of water at $50\ ^\circ\text{C}$ is poured into a vessel containing $120\text{ g}$ of water at $10\ ^\circ\text{C}$. The final temperature recorded is $20\ ^\circ\text{C}$. Calculate the thermal capacity of the vessel. Consider that the vessel is at the same initial temperature as the cold water.

We have

$\text{Substance}$ $\hspace{10mm} \text{Mass} \hspace{10mm}$ $\hspace{10mm} \text {S.H.C} \hspace{10mm}$ $\hspace{5mm} \text{Initial temperature} \hspace{5mm}$ $\hspace{10mm} \text{Change in temperature}\hspace{10mm}$ $\text{Hot water}$ $100\text{ g}$ $4.2\text{ J/g}\, ^\circ\text{C}$ $50\ ^\circ\text{C}$ $(50 - 20)\ ^\circ\text{C} = 30\ ^\circ\text{C}$ $\text{Cold water}$ $120\text{ g}$ $4.2\text{ J/g}\, ^\circ\text{C}$ $10\ ^\circ\text{C}$ $(20 - 10)\ ^\circ\text{C} = 10\ ^\circ\text{C}$ $\text{Vessel}$ $?$ $?$ $10\ ^\circ\text{C}$ $(20 - 10)\ ^\circ\text{C} = 10\ ^\circ\text{C}$ $\begin{aligned} (\text{The thermal capacity of the vessel}) & = m × c = x (\text{say})\\ \\ (\text{Heat lost by hot water}) & = m × c × {\theta}_F\\ & = 100 × 4.2 × 30\\ & = 12600\text{ J}\\ \\ (\text{Heat gained by cold water}) & = m × c × {\theta}_R\\ & = 120 × 4.2 × 10\\ & = 5040\text{ J}\\ \\ (\text{Heat gained by the vessel}) & = m × c × {\theta}_R\\ & = x × 10 (\text{substituting the value of }mc = x)\\ & = 10x \text{ J}. \end{aligned}$

According to the law of conservation of energy,

$\begin{aligned} (\text{Heat gained}) & = (\text{Heat lost})\\ 5040 + 10x & = 12600\\ 10x & = 7560\\ \Rightarrow x & = 756. \end{aligned}$

Therefore, the thermal capacity of the vessel is$756\text{ J/}^\circ\text{C}. \ _\square$

## Applications

There are quite a few applications of heat transfer between particles. We will be discussing two of them.

Green House Effect

Thermos Flask