As the density of water increases and volume of water decreases from a °C to 4°C, so the volume x of the wooden block will increase till the temperature of water becomes 4°C. Now, as the temperature increases from 4°C to 8°C, the density of water decreases and its volume increases above 4°C, therefore the volume x of the block will also decrease.

Suppose the temperature of the water in the smaller vessel is 8 at time t. In the next time interval dt, a heat, \(\triangle\)\(\theta\) is transferred to it where
\(\triangle \theta =\frac { KA }{ L } \left( { \theta }_{ 0 }-{ \theta } \right) dt\)
This heat increases the temperature of the water of mass 'm' to \(\theta\) + d\(\theta\)
Where  \(\triangle\)\(\theta\) = ms d\(\theta\)
From eqn. (i) and (ii),
\(\frac { KA }{ L } \left( { \theta }_{ 0 }-{ \theta } \right) dt=msd\theta \)
or dt = \(\frac { Lms }{ KA } \frac { d\theta }{ { \theta }_{ 0 }-\theta } \)
\(\Rightarrow\) \(\int _{ 0 }^{ T }{ dt } =\frac { Lms }{ KA } \int _{ { \theta }_{ 1 } }^{ { \theta }_{ 2 } }{ \frac { d\theta }{ { \theta }_{ o }-\theta } } \)
where T is the time required for the temperature of the water to become \(\theta\)₂
Thus, \(\left[ \frac { Lms }{ KA } In\frac { { \theta }_{ 0 }-{ \theta }_{ 1 } }{ { \theta }_{ 0 }-{ \theta }_{ 2 } } \right] \)

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)

\(\frac { \triangle \theta _{ 1 } }{ \triangle t } =\frac { KA\left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ x } \)
The rate of heat generation in the box = 13 W. The rate at which heat flows out of the box through the right plug is
\(\frac { \triangle \theta _{ 1 } }{ \triangle t } +13W=\frac { \triangle \theta _{ 2 } }{ \triangle t } \)
or \(\frac { KA }{ x } \left( { \theta }_{ 1 }-\theta \right) +13W=\frac { KA }{ x } \left( { \theta }-\theta _{ 2 } \right) \)
or \(2\frac { KA }{ x } \theta =\frac { KA }{ x } \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) +13W\quad \Rightarrow \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } +\frac { (13w)x }{ 2KA } \)
or \(\theta =\frac { { 100 }^{ 0 }C+4C }{ 2 } +\frac { (13w)x0.08 }{ 2x(2.0W/m-°C)\left( 12\times { 10 }^{ -4 }{ m }^{ 2 } \right) } \)
= 52°C + 216.67°C \(\equiv \) 269°C.

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 = \(\sigma T^{ 4 }\)
\(\\ T^{ 4 }=\frac { E }{ \sigma } \Rightarrow T=\left( \frac { E }{ \sigma } \right) ^{ { 1 }/{ 4 } }=\left( \frac { 5.67\times 10^{ 4 } }{ 5.67\times 10^{ -8 } } \right) ^{ { 1 }/{ 4 } }\)
\(\\ T=(10^{ 12 })^{ { 1 }/{ 4 } }=10^{ 3 }\)
\(\\ T=1000K\)

(i) -148^o F
(ii) 32^oZ

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

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
\( \Rightarrow \Delta T=\frac{g l \sin \theta}{5}=\frac{10 \times \frac{60}{100} \times \frac{3}{5}}{420}\)
\( =8.6 \times 10^{-3}{ }^{\circ} \mathrm{C} \)