Thermal Physics

10 methods · 16 questions

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Every mode of heat transfer responds to the DIFFERENCE between two temperatures, so shifting both by the same amount changes nothing, and raising the colder one actually slows the transfer. Examiners build whole option lists out of candidates who track $T_{1}$ and $T_{2}$ separately.

Trigger: an option list in which the two temperatures are moved up or down independently, or a numbered list offering 'increase $T_{1}$ ' and 'increase $T_{2}$ ' as separate changes.

Instances:
(i) both temperatures shifted the same way by the same amount: the difference, and so the rate, is unchanged;
(ii) raising the COLD end reduces the difference and slows the transfer, even though a temperature has gone up;
(iii) equal and opposite shifts change the difference by twice the shift, not by the shift;
(iv) a body sitting at exactly the temperature of the fluid around it has zero difference, so it drives no convection at all;
(v) any option built on a temperature SUM such as $T_{1} + T_{2}$ is unphysical and can be struck out on sight.

Linked questions (3)

NSAA 2016 Section 1 Part B (ESAT Physics) · Q21

A cylindrical copper bar X of length

l

has a cross-sectional area

A

. The colder end of the bar is kept at temperature

T_{1}

and the hotter end is kept at temperature

T_{2}

.

The curved surface of the bar is perfectly insulated and thermal energy is conducted from the hotter end of the bar to the colder end.

Changes in

l

A

T_{1}

and

T_{2}

may alter the rate at which thermal energy is conducted along the copper bar.

In which case is the rate of conduction of thermal energy along the insulated bar the same as for bar X?

A $A$ doubles; $l$ , $T_{1}$ , $T_{2}$ unchanged
B $A$ halves; $l$ , $T_{1}$ , $T_{2}$ unchanged
C $l$ doubles; $A$ , $T_{1}$ , $T_{2}$ unchanged
D $l$ halves; $A$ , $T_{1}$ , $T_{2}$ unchanged
E $A$ , $l$ unchanged; $T_{1}$ decreases by $10^{\circ} C$ , $T_{2}$ increases by $10^{\circ} C$
F $A$ , $l$ unchanged; $T_{1}$ decreases by $10^{\circ} C$ , $T_{2}$ unchanged
G $A$ , $l$ unchanged; $T_{1}$ increases by $10^{\circ} C$ , $T_{2}$ decreases by $10^{\circ} C$
H $A$ , $l$ unchanged; $T_{1}$ increases by $10^{\circ} C$ , $T_{2}$ increases by $10^{\circ} C$

Worked solution

Answer:H

The rate of conduction along an insulated bar is

P = \frac{k A ( T _{2} - T _{1} )}{l}

, so it depends on

A

, on

l

, and on the temperature difference

(T_{2} - T_{1})

, never on

T_{1}

and

T_{2}

individually.

Options A to D each change

A

l

alone, which changes

P

directly (doubling

A

doubles

P

; doubling

l

halves

P

), so none of them leaves the rate unchanged.

Options E to H each shift both

T_{1}

and

T_{2}

10^{\circ} C

, leaving

A

and

l

unchanged, so only the change in

(T_{2} - T_{1})

matters:
- E:

T_{1}

down

10

T_{2}

10

, so

(T_{2} - T_{1})

increases by

20^{\circ} C

.
- F:

T_{1}

down

10

T_{2}

unchanged, so

(T_{2} - T_{1})

increases by

10^{\circ} C

.
- G:

T_{1}

10

T_{2}

down

10

, so

(T_{2} - T_{1})

decreases by

20^{\circ} C

.
- H:

T_{1}

10

T_{2}

10

, so

(T_{2} - T_{1})

is unchanged.

Only in H does the temperature difference stay the same, so only H leaves the rate of conduction unchanged.

The answer is H.

NSAA 2016 Section 1 Part B (ESAT Physics) · Q31

The diagram shows four solid steel balls P, Q, R and S which are of identical size.

Balls P and R have shiny surfaces. Balls Q and S have dull surfaces.

Balls P and Q are in a room at

20^{\circ} C

. Balls R and S are in a room at

40^{\circ} C

.

At a given moment, ball P is at

30^{\circ} C

, ball Q is at

30^{\circ} C

, ball R is at

40^{\circ} C

and ball S is at

40^{\circ} C

.

Which two balls lose thermal energy by convection, and which ball emits thermal radiation at the greatest rate?

Two rooms are shown side by side. The left room, labelled 'room at 20 degrees C', contains two identical circles: one drawn with a plain white (shiny) outline, labelled P and marked 30 degrees C, and one drawn filled solid black (dull), labelled Q and also marked 30 degrees C. The right room, labelled 'room at 40 degrees C', contains two more identical circles in the same style: a plain white (shiny) one labelled R, marked 40 degrees C, and a filled black (dull) one labelled S, also marked 40 degrees C. — illustrative

AP and Q; P
BP and Q; Q
CP and Q; R
DP and Q; S
ER and S; P
FR and S; Q
GR and S; R
HR and S; S

Worked solution

Answer:D

Convection is heat transfer by the bulk movement of a fluid (here, the air in each room), driven by a temperature difference between an object and the air around it: air warmed by a hotter object expands, becomes less dense, and rises, setting up a convection current. Balls P and Q are at

30^{\circ} C

in air at

20^{\circ} C

, a

10^{\circ} C

difference, so both drive convection in their room. Balls R and S are at

40^{\circ} C

in air also at

40^{\circ} C

: no temperature difference, so neither drives any convection (each is in thermal equilibrium with its room). So P and Q are the balls losing thermal energy by convection.

Every object continuously emits thermal radiation, at a rate that increases with its surface temperature and with its emissivity: a dull, dark surface emits (and absorbs) radiation much more readily than a shiny, reflective one at the same temperature. Comparing the four balls, S is both dull (the higher-emissivity finish) and at the higher temperature (

40^{\circ} C

, versus

30^{\circ} C

for the two balls at

20^{\circ} C

room temperature), and no other ball combines as high a temperature with as high an emissivity. So S emits thermal radiation at the greatest rate. (This is about the rate of emission itself, not the net transfer: R and S being in equilibrium with their room means no net heat flows for them, but each still radiates, and absorbs, continuously.)

The answer is D.

NSAA 2018 Section 1 Part B (ESAT Physics) · Q24

A student is investigating heat flow along a solid uniform metal bar.

The bar has length

l

, cross-sectional area

A

, and has its ends maintained at temperatures

T_{1}

and

T_{2}

(where

T_{1} > T_{2}

).

Which relationship represents the rate of heat flow

P

along the bar?

(Assume that there is no heat transfer through the sides of the bar.)

A $P \propto \frac{( T _{1} - T _{2} )}{A l}$
B $P \propto \frac{( T _{1} + T _{2} )}{A l}$
C $P \propto \frac{A ( T _{1} - T _{2} )}{l}$
D $P \propto \frac{A ( T _{1} + T _{2} )}{l}$
E $P \propto \frac{l}{A ( T _{1} - T _{2} )}$
F $P \propto \frac{l}{A ( T _{1} + T _{2} )}$
G $P \propto \frac{A l}{T _{1} - T _{2}}$
H $P \propto \frac{A l}{T _{1} + T _{2}}$

Worked solution

Answer:C

For thermal conduction along a uniform bar (Fourier's law of conduction), the rate of heat flow is directly proportional to the cross-sectional area

A

(a wider bar conducts more heat) and to the temperature difference

(T_{1} - T_{2})

driving the flow, and inversely proportional to the length

l

(a longer bar conducts less, for the same temperature difference, since the temperature gradient is shallower):

P \propto \frac{A ( T _{1} - T _{2} )}{l} .

The answer is C.