Temperature, Heat, and the first law of thermodynamics formulas

Question 1

Temperature (constant-volume gas thermometer)

Answer 1

T = (273.16K) × lim gas -> 0 (p/p3)

Question 2

Conversion of Other Temps. from Kelvin temp

Answer 2

Celsius Temp = T - 273.15

Farenheit Temp = 9/5 Celsius temp + 32

Question 3

Thermal expansion (change in L)

Answer 3

ΔL = LαΔT

α is the coefficient of linear expansion

Question 4

Thermal expansion (change in volume)

Answer 4

ΔV = VβΔT

β = 3α is the coefficient of volume expansion

Question 5

Heat conversions (calories, joules. british thermal units)

Answer 5

1 cal = 3.968 × 10 ^(-3) Btu = 4.1868 J

Question 6

relationship of temperature change with heat Q

Answer 6

Q = C (Tf - Ti)

C is the heat capacity of the object

Question 7

Heat of an object when it has mass

Answer 7

Q = cm (Tf - Ti)

c is the specific heat of the material
molar specific heat = 6.02 × 10^(23)

Question 8

Heat transformation

Answer 8

Q = Lm

Question 9

Work Associated with volume Change

Answer 9

W = ∫dW = ∫p dV [ from Vi to Vf]

Question 10

First Law of Thermodynamics

Answer 10

ΔE int = Eint,f - Eint,i = Q - W

dEint = dQ -dW

Question 11

Application of the First Law

Answer 11

adiabatic processes:
Q = 0, ΔEint = -W

constant-volume processes:
W = 0, ΔEint = Q

cyclical processes:
ΔEint = 0, Q = W

free expansions:
Q = W = ΔEint = 0

Question 12

Conduction of Energy (Rate)

Answer 12

Pcond = Q/t = kA [(TH - TC)/L]

TH = higher temp
TC = lower temp

Question 13

Rate of radiation energy

Answer 13

Prad = σεAT^4

σ = 5.6704 × 10 ^(-8) W/m^2×K^4
ε is the emissivity of the object’s surface

Question 14

Stefan-Boltzmann constant

Answer 14

σ = 5.6704 × 10 ^(-8) W/m^2×K^4

Question 15

Rate of radiation energy absor[topm

Answer 15

Pabs = σε AT^4(env)