thermal properties of matter

(reading on any scale -lower fixed point (ice point))/(upper fixed point(steam point)-lower fixed point) =constant i.e C/5=(F-32)/9=(K-273.15)/5

C/5= F-32/9 ∆F=(9/5)∆C, ∆K=∆C

∆l=l*(alpha)*∆T, ∆A=A*(beta)*∆T, ∆V = V*(gamma)*∆T

beta =2*alpha, gamma=3*alpha, ratio alpha:beta:gamma= 1:2:3

L1*(alpha1)=L2*(aplha2)

aplha(series)=(L1*aplha1+L2*aplha2)/L1+L2 , alpha(parallel)=(((aplha1)/L1)+(aplha2/L2))/((1/L1)+(1/L2))

T=2π√(l°(1+aplha*∆t) here ∆t is temperature difference l° is initial length,∆T/T=aplha*∆t/2 ,%timenlostor gain =(aplha*∆t/2) *100 , time lost or gain per day ∆T=43200*(alpha)*∆t

invar =64%iron + 36% nickel (coefficient of linear expansion is very small for invar pendulum are made from invar to show correct time in all seasons)

TRUE VALUE=SR(1+aplha*∆T) ,SR: scale reading,here body do not expand ,due to linear expansion of scale the scale will expand and scale reading will be less than the true value at T' (if a scale gives correct reading at temperature T), however if T'<T due to contraction scale reading will be more than the true value ,so true value will still be given by the above equation ∆T=(T'-T) negative

strain =∆l/l°=alpha*∆T , young's modulus Y=Fl°/A*∆l

· Bimetallic strip works on the principle of differential expansion of metals
· If a bimetallic strip made of brass and iron is heated, brass bends on convex side (\alpha_{\text{brass}} > \alpha_{\text{Fe}})
· If it is cooled brass bends on concave side. The reason for this is expansion/ compression of brass is more than that of iron. Radius of curvature of bimetallic strip R=t/((alphabrass-alphaFe)∆T) where t is half the thickness of bimetallic strip used

Thermal expansion of an isotropic object may be imagined as a photographic enlargement. So if there is a hole A in a plate C (or cavity A inside a body C), the area of hole (or volume of cavity) will increase when body expands on heating, just as if the hole (or cavity) were solid B of the same material. Also the expansion of area (or volume) of the body C will be independent of shape and size of hole (or cavity), i.e., will be equal to that of D

(1) Liquid do not have linear and superficial expansion but these only have volume expansion
(2) Since liquids are always to be heated along with a vessel which contains them, so initially on heating the system (liquid + vessel), the level of liquid in vessel falls(as vessel expands more since it absorbs heat and liquid expands less) but later on, it starts rising due to faster expansion of the liquid

EXPANSION OF LIOUID
Expansion liquid is called real expansion.
Expansion of solid is called expansion of vessel
Observed expansion is called apparent expansion. gamma(real)=∆V(real)/V∆T. gamma(vessel)=∆Vvessel/V∆T. ∆Vapp=V∆T[gamma(real)-gamma(vessel)] also gamma(app)=gamma(real)-gamma(vessel)

amma \Delta V Level
amma_{\text{real}} > amma_{\text{vessel}} (= 3\alpha) \Rightarrow amma_{\text{app}} > 0 \Delta V_{\text{app}} is positive Level of liquid in vessel will rise on heating
amma_{\text{real}} < amma_{\text{vessel}} (= 3\alpha) \Rightarrow amma_{\text{app}} < 0 \Delta V_{\text{app}} is negative Level of liquid in vessel will fall on heating
amma_{\text{real}} = amma_{\text{vessel}} (= 3\alpha) \Rightarrow amma_{\text{app}} = 0 \Delta V_{\text{app}} = 0 Level of liquid in vessel will remain same(look at DeepSeek)

p'=p/(1+gamma*∆T) (for small temperature change p'=p(1-gamma*∆T)

Generally matter expands on heating and contracts on cooling. In case of water, it expands on heating if its temperature is greater than 4^{\circ}\text{C}. In the range 0^{\circ}\text{C} to 4^{\circ}\text{C}, water contracts on heating and expands on cooling. i.e amma is negative. This behaviour of water in the range from 0^{\circ}\text{C} to 4^{\circ}\text{C} is called anomalous expansion.

1 cal 4.2J

s=∆Q/M∆T, ∆Q=integral from T1 to T2[M*s*dT]

s=∆Q/∆T

∆Qwater=∆Qbody ,water equivalent=mass of body ×specific heat of the material

Teq=(m1s1T1+m2s2T2)/(m1s1+m2s2)

solid to liquid (melting), liquid to solid (freezing), liquid to vapour(vaporisation),vapour to liquid (condensation) , solid to vapour (sublimation) Liquid to vapour (evaporation, without heating) melting happens at 0°C boiling happens at 100°C

Boiling point increases with increase in pressure
If you close the steam outlet pressure increases ightarrow boiling stops. So more heat is required to boil again thus boiling point increases This is why pressure cooker cooks faster (BP of water > 100°C inside)
Similarly Boiling point decreases with decrease in pressure

The specific temperature and pressure at which all three phases of water - Solid (Ice) liquid (water), and gas (Water vapor) - coexist simultaneously in thermodynamic equilibrium is called triple point.

L=∆Q/m ,where m is in kg

Lf=80cal or 3.36*10^5 Jkg^-1

Lv= 540 cal or 2.26*10^6 Jkg^-1

power =∆Q/∆t , ms(ice)∆T/∆t1 =mLf/∆t2 =ms(water)∆T/∆t3. =mLv/∆t4

∆Q/∆t=KA(T1-T2)/l =-KA∆T/l where T1>T2 and ∆T = T2-T1 ,in differential form dQ/dt=-KAdT/dx, ∆Q/∆t Is called average heat current (H) H=-∆T/Rth (Rth=l/KA),unit of Rth is kelvin/Watt , dT/dx is called temperature gradient,heat flows in the direction of decreasing temperature

theta(i)=(theta1/R1+theta2/R2 +theta3/R3)/(1/R1+1/R2+1/R3) at study state rate of flow of heat through the conductor is constant

H is constant. Keq(series)=(L1+L2+l3)/((L1/k1)+(L2/k2)+(l3/k3)) , Keqparallel (length is same)K=(K1A1+K2A2+...+KnAn)/(A1+A2+...+An)

E=Energy radiated/(area×time)

E=energy radiated/(Time*area*wavelength)=dQ/(dt*A*d(lamda) , total emissive power (E)=intergralfrim zero to infinity[E d(lamda)]

e=energy radiated by a body /energy radiated by ideal black body (ranges from 0<e<1 ,if e=0 no energy is radiated if e=1 then that body is perfect black body)

E=sigmaT^4 ,where sigma is stefan's constant having dimension[MT^3theta^-4] and value is 5.67*10^-8 Wm^-2K^-4 for normal body E=e*sigma*T^4,Q=A*e*sigma*T^4*t , radiant power P=Q/t=A*e*sigma*T^4 P is proportional to R^2T^4,

E=e*sigma(T^4-T°^4) ,for black body E=sigma(T^4-T°^4) note: good absorber is good emitter if the object is in thermal in equilibrium)

dT/dt= e*A*sigma(T^4-T°^4)/mc (where c is specific heat)

lamda*T=b b is weins constant value is 2.89×10^-3 mK ,area under graph gives power of boder therefore A1/A2=[T1/T2]^4 ,height of peak is proportional to T^5

when T-T°>or =35°C, -dT/dt =K(T-T°)

ln[Tf-T°]-ln[Ti-T°] or. ∆Tf=∆Ti e^-Kt

T1-T2/t =K[(T1+T2)/2 -T°]