For thermal expansion and its classification, see text.
Let V cm³' be the volume of either liquid. Then the thermal capacity of each calorimeter
is also V cal per ⁰c.
Mass of water = V x 1 = V g
Mass of alcohol = V x 0.8 = 0.8 V g
The rate of cooling of the 'water calorimeter'
= \(\frac { 1 }{ 100 } \) IV x (50 - 40) + V x 1 x (50- 40)}
= 5 V CalS^-1
= The rate of cooling of the' alcohol calorimeter' 
= \(\frac { 1 }{ 74 } \) (V x (50-40) +0.8 V x s x(50 - 40)}
= \(\frac { 1 }{ 74 } \) (10V + 8Vs)
Because identical volumes of the liquids are getting cooled under identical conditions, the rate of cooling is the same in both the cases. Hence
5V = \(\frac { 1 }{ 74 } \) (10 V + 8 Vs) which gives
s = 0.6 cal g^-1 (⁰C^-1)

Given, T =1327 + 273 = 1600K
Wien's constant , b = 2.9 x 10^-3 m K
\(\lambda _{ m }=\frac { b }{ t } =\frac { 2.9\times 10^{ -3 } }{ 1600 } =1.81X10^{ -6 }m\)

Given, \(\sigma \) = 5.67W / cm^-2 =5.67 x 10⁴ W/m²
\(\sigma \) = 5.67 x 10^-8 Wm^-2 K^-4
Apply Stefan's law, E = \(\sigma T^{ 4 }\)
\(\\ T^{ 4 }=\frac { E }{ \sigma } \Rightarrow T=\left( \frac { E }{ \sigma } \right) ^{ { 1 }/{ 4 } }=\left( \frac { 5.67\times 10^{ 4 } }{ 5.67\times 10^{ -8 } } \right) ^{ { 1 }/{ 4 } }\)
\(\\ T=(10^{ 12 })^{ { 1 }/{ 4 } }=10^{ 3 }\)
\(\\ T=1000K\)

8.6 × 10^∘C

As block slides along x only mgsinθ and friction F do work, let them be mglsin θ and wF.
Now, speed is constant
∴ All the loss in potential energy is dissipated as heat, no. kinetic energy being gained.
(s is the specific heat) ⇒ mglsinθ = wF = Heat lost = ms Δ T
\( \Rightarrow \Delta T=\frac{g l \sin \theta}{5}=\frac{10 \times \frac{60}{100} \times \frac{3}{5}}{420}\)
\( =8.6 \times 10^{-3}{ }^{\circ} \mathrm{C} \)

If Q be the amount of heat flowing from hot to the cold face, then it is found to be:
(i) directly proportional to the cross-sectional area (A) of the face.
i..e   Q \(\infty\) A
(ii) directly proportional to the temperature difference between the two faces i.e.,
i..e  Q \(\infty\) \(\triangle\)\(\theta\)
(iii) directly proportional to the time t for which the heat flows
i..e., Q \(\infty\) t
(iv) inversely proportional to the distance' d' between the two faces
i..e Q \(\frac { 1 }{ \triangle x } \)
Combining factors (1) to (4), we get
\(Q\infty \frac { A\triangle \theta }{ \triangle x } t\)
or \(Q=KA\frac { \triangle \theta }{ \triangle x } t\)
where K is the proportionality constant known as the coefficient of thermal conductivity

We know that Q = \(\frac { KA\left( { T }_{ 1 }-{ T }_{ 2 } \right) t }{ l } \)
For given problem , kt = constant or K \(\infty \) \(\frac { 1 }{ t } \)
\(\therefore\)  \(\frac { { k }_{ 1 } }{ { k }_{ 2 } } =\frac { { t }_{ 2 } }{ { t }_{ 1 } } =\frac { 10 }{ 30 } =\frac { 1 }{ 3 } \)

The length of 1 em division of the steel scale at 40°C is
(1 cm) x (1 + 11 x 10-6 x 20) = 1.00022 cm
Length of the copper rod at 40°C will be (80\ x (1 + 17 x 10^-6 x 20) = 80.0272 cm. The number of em read on the scale will be
\(\frac { 80.0272 }{ 1.00022 } \) cm = 80.0096 cm

1.35 x 10⁴ W

\(11\times { 10 }^{ 5 }N\)

Heat lost by hot body = Heat gained by cold body

Infinite, because \(\triangle\)T = 0; Use C = \(\frac { \triangle Q }{ m\triangle T } \)

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

Let be the same numerical value of temperature on the both scales.
\(\frac { T_{ C } }{ 5 } =\frac { T_{ F }-32 }{ 9 } \) 
\(\\ \Rightarrow \frac { \theta }{ 5 } =\frac { \theta -32 }{ 9 } [\because \theta ^{ 0 }C=\theta ^{ 0 }F=\theta ^{ 0 },\ given]\)
\(\\ \Rightarrow 9\theta =5\theta -160\)
\(\\ -4\theta =160\)
\(\\ \therefore \ \theta =-40^{ 0 }\)
\(\\ \theta =-40^{ 0 }C=-40^{ 0 }F\) 

As both flasks have same height and capacity, the area of the cylindrical wall will be less than that of the square wall. Hence, the thermos flask of circular cross section will transmit less heat as compared to the thermos flask of square cross section and will be better.

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.

This is because metals are good conductors of heat and have low specific heat capacity.

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.

The two bodies may have different masses and different materials i.e., they may have different thermal capacities. In case the two bodies have equal thermal capacities, they would settle at the mean \(\frac { ({ T }_{ 1 }+{ T }_{ 2 }) }{ 2 } \)

t(⁰C) = T(k) - 273.15 or
T(k) = t(⁰C) + 273.15

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

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.