\(For\ brass\ rod,\ l=50\ cm,\ t_{ 1 }=40^{ 0 }C,\ t_{ 2 }=250^{ 0 }C\)
\(\ \alpha =2.0\times 10^{ -5 }\ ^{ 0 }C^{ -1 }\)
\(\\ Change\ in\ length\ of\ brass\ rod\ is\)
\(\\ \triangle l=\alpha l({ t }_{ 2 }-{ t }_{ 1 })\)
\(\\ =2.0\times 10^{ -5 }\times 50\times (250-40)=0.21cm\)
\(\\ For\ steel\ rod,\ l=50cm,t_{ 1 }=40^{ 0 }C,\ t_{ 2 }=250^{ 0 }C,\)
\(\\ \ \alpha =1.2\times 10^{ -5 }\ ^{ 0 }C^{ -1 }\)
\(\\ Change\in\ length\ o f\ steel\ rod\ is\)
\(\\ \triangle l'=\alpha l({ t }_{ 2 }-{ t }_{ 1 })\)
\(\\ =1.2\times 10^{ -5 }\times 50\times (250-40)=0.13cm\)
\(\\ Change\ in\ length\ of\ the\ combined\ rod\ at\ 250^{ 0 }C\)
\(\\ =\triangle l+\triangle l'=0.21+0.13=0.34cm\)
\(\\ =1.2\times 10^{ -5 }\times 50\times (250-40)=0.13cm\)
\(\\ Change\ in\ length\ of\ the\ combined\ rod\ at\ 250^{ 0 }C\)
\(\\ =\triangle l+\triangle l'=0.21+0.13=0.34cm\)
As the rods expand freely, so no thermal stress is developed at the junction.

When T=273K, =101.6Ω
∴ 101.6=R0[1+5×10⁻³(273−T₀)]
Given, T=600.5K, R=165.5Ω
∴ 165.5=R0[1+5×10⁻³(600.5−T₀)]
Dividing Eq. (ii) by Eq(i) we get
\(\\ \frac { 165.5 }{ 101.0 } =\frac { 1+5\times 10^{ -3 }(600.5-T_{ 0 }) }{ 1+5\times 10^{ -5 }(273-T_{ 0 }) } \)
on solving, T₀=−49.3K
Substituting in Eq. (i) we get
\(\\ 101.6=R_{ 0 }[1+5\times 10^{ -3 }(373+-49.3)]\)
\(\\ or\ { R }_{ 0 }=\frac { 101.6 }{ 1+5\times 10^{ -3 }\times 322.3 } =38.9\Omega\)
For R=123.4 Ω, we have
\(\\ 123.4=38.9[1+5\times 10^{ -3 }(T+49.3)]\)
On solving, we get T=384.8 K

Explanation,
1. There are two important temperature scales, Celsius and Fahrenheit used in the world.
2. There is a point on both scales where the temperatures in degrees are equal.
3. This is -40° and -40°.
Celsius and Fahrenheit,
Celsius is directly proportional to Fahrenheit.
1. This says that with the increase in the temperature on the Celsius scale, its Fahrenheit temperature equivalent temperature will also rise.
2. When the temperature on the Celsius scale decreases, its Fahrenheit temperature equivalent temperature will also be low.
3. Multiply the °C temperature by 1.8 Add 32 to this number. This is the answer in °F
4. The Celsius freezing point of water is 0°C and its boiling point is 100°C
5. 100°C-0°C=100°C between water's freezing and boiling points.
Conversion of Celsius to Fahrenheit is F = 9/5 C + 32
Conversion from Fahrenheit to Celsius is C = \(\frac{5}{9}(F-32)\)
Step 1: Given
Both celcius and fahrenheit have same temperature.
Let the temperature = x
Step 2: Formula used.
The formula as given is C/100 = \(\frac{(F-32)}{180}\)
Step 3: Calculation
We want to know the point at which the temperatures are the same, so let's call that temperature 
\( \text { or, } \frac{x}{100}=\frac{(x-32)}{(180)} \)
\( \text { or, } x=\frac{100 *(x-32)}{180} \)
\(\text { or, } x=\frac{50 *(x-32)}{90} \)
\(\text { or, } x=\frac{5 \times(x-32)}{9} \)
\(\text { or, } x=\frac{5 x-5 \times 32}{9} \)
\(\text { or, } x=\frac{5 x}{9}-\frac{(5 \times 32)}{9} \)
\(\text { or, } x=\frac{5 x}{9}-\frac{160}{9} \)
\(\text { or, } 9 x=5 x-160\)
\(\text { or, } 9 x-5 x=-160 \)
\(4 x=-160 \)
\(x=-40 \)
Hence, at -40°C and -40°C temperature are Celsius and Fahrenheit equal.

Here m=0.5 m²=500L=500 kg
So the heat liberated during the water changes 20^∘C to 5^∘C
=500×4200×15
[Δθ=20−5=15]
=500×4200×15
=75×420×1000
=31500×1000
Let the height = h
the required work
=mgh=10×10×h=100 h
But, 100 h = 3150000
⇒h=315000 m=315 km

A cooking utensil should be of suitable material, which possesses small valve of specific heat (C), large value of thermal conductivity (K) and low coefficient of expansion \((\alpha )\)

Heat carried away from a fire sideways mainly by radiation. Above the fire, heat is carried by both radiation and convection of air. But convection carries much more heat than radiation. So, it is much hotter above a fire than by its sides

A cooking utensil should have
(i) high conductivity so that it can conduct heat through itself and transfer it to the contents quickly.
(ii) low specific heat so that it immediately attains the temperature of the source.

The object at 300^oC will cool faster than the object at 100^oC. This is in accordance with Newton’s law of cooling. As we know, rate cooling of an object α  temperature between the object and its surroundings.

The neck expands but not the stopper due to poor conductivity of glass. Thus, the stopper can be taken out easily.

The area of the faces will increased by 0.02% and the volume by 0.03%.

yes, water at 30^oC can be converted into vapour by reducing its pressure until it equals to the vapour pressure of water at 30^oC.

The thermometer bulb having small heat capacity will absorb less heat from the body whose temperature is to be measured. Hence, the temperature of that body will practically remain unchanged.

Electrons contribute largely both towards the flow of electricity and the flow of heat. A good conductor contains a large number of free electrons. So, it is both a good conductor of heat and electricity.

When the birds swell their feathers, they are able to enclose air in the feathers. Air, being a poor conductor of heat, so it prevents the loss of heat from the bodies of the birds to the surroundings and as such they do not feel cold in winter.

\(\frac{C-0}{100-0}=\frac{F-32}{212-32}=\frac{R-0}{80-0}\)

No, the process of transfer of heat by convection is based on the fact that a liquid becomes lighter .on becoming hot and rise up. In condition of weightlessness, this is not possible. So, transfer of heat by convection is not possible in the earth satellite.

The statement is true. A black body absorbs radiations of all wavelengths. When heated to a suitable temperature, it emits radiations of all wavelengths. Hence, a black body radiation is white.

-5.8 kg/ m³

Given, cross-section area, A=2.0 mm²=2x10^-6m²
change in temperature, \(\triangle T=30-5=25^{ 0 }C\)
Young's modulus of steel wire, Y = 2 x 10¹¹Nm^-2
and coefficient of linear expansion of steel,
\(\alpha =1.1\times 10^{ -5 }/^{ 0 }C\)
Tension developed in the rod,
\(F=YA\alpha \triangle T\)
\(\\ =2\times 10^{ 11 }\times 2\times 10^{ -6 }\times 1.1\times 10^{ -5 }\times 25\)
\(\\ F=110N\)

As the brass scale of a barometer gives correct reading at T₁ = 0⁰ C , hence at temperature T₂ = 27⁰ C., the scale will expand and will not give correct reading
In such, true value
= Observed scale reading x\((1+\alpha \triangle T)\)
True pressure = 75.00 cm x [1 + 2.0 x 10^-5 x (27 - 0)]
= 75 x (1 + 2.0 x 10^-5 x 27]
= 75.00(1 + 54 x 10^-5) cm = 75.04 cm

Here, A = 6 x  Side² = 6 x 30 x 30
= 5400cm² = 0.54m²
x = 5 cm = 0.05 m
t = 6 h = 6 x 3600s
T₁-T₂ = 45-0 = 45⁰C,
K = 0.01Js^-1m^-10C^-1
L = 335 x 10³ Jkg^-1
Total heat entering the box through all the six faces,
Q = \(\frac { KA(T_{ 1 }-T_{ 2 })t }{ x }\)
\( \\ =\frac { 0.01\times 0.54\times 45\times 6\times 3600 }{ 0.05 } =104976J\)
Let m kg of ice melt due to this heat. Then,
Q = mL
OR
\(m=\frac { Q }{ L } =\frac { 104976J }{ 336\times 10^{ 3 }Jkg^{ -1 } } =0.313kg\)
Mass of ice left after six hours = 4 - 0.313 = 3.687 Kg

Here, A= 0.15 m², x = 1.0 cm = 0.01 m,
K = 109 Js^-1 m^{-1\(^{0}C-1, \)} L= 2256 Jg^-1
T₂ = 100\(^{0}C, \) t = 1 min = 60 s
Let T₁ be the temperature of the part of the flame in contact with boiler. Then, amount of heat that flows into water in 1 min.
\(Q = \frac { KA({ T }_{ 1 }-{ T }_{ 2 })t }{ x } =\frac { 109\times 0.15\times ({ T }_{ 1 }-100)\times 60 }{ 0.01 } J\)
Mass of water boiled per min = 6 kg = 6000 g
Heat used to boil water,
Q = mL= 6000 g x 2256 Jg^-1 = 6000 x 2256 J
\(\therefore \frac { 109\times 1.15\times ({ T }_{ 1 }-100)\times 60 }{ 0.01 } =6000\times 2256\)
\(or { T }_{ 1 }-100=\frac { 6000\times 2256\times 0.01 }{ 109\times 0.15\times 60 } =138^{ 0 }C\)
or T₁ = 138 + 100
= 238\(^{0}C\) 

0.42 W m^-1K^-1