In the diagram below, metal 1 is made of copper and metal 2 is silver. Both bars
ID: 1631692 • Letter: I
Question
In the diagram below, metal 1 is made of copper and metal 2 is silver. Both bars have the same length and area of cross-section but their thermal conductivities are different. The temperature Th is maintained at a constant 82.0°C at the end A while the temperature Tc at the end B is maintained at 26.0°C.
(a) What is the temperature Tj at the junction of the two bars? The thermal conductivity of copper is 390 J/(s · m · °C) and that for silver is 420 J/(s · m · °C).
_____°C
(b) If metals 1 and 2 are now interchanged, but the values of Th and Tc are the same as in part (a), what will be the temperature of the junction?
____°C
Question 2
Suppose you stand with one foot on ceramic flooring and one on a wool carpet, making contact over an area of 84.0 cm2 with each foot. Both the ceramic and the carpet are 1.50 cm thick and are 10.0°C on their bottoms. At what rate in watts must each foot supply heat to keep the top of the ceramic and carpet at 33.0°C?
wool carpet _____ W
ceramic tile _____ W
PLEASE ANSWER ALL THE QUESTIONS DONT GIVE ANY WRONG ANSWERS THANKS FOR YOUR HELP !
Explanation / Answer
(a)
let junction temperature be Tj
let area of each cross section is Ac and length of each part is L
then asuming there is no heat loss to the environment,
heat conducted from A to the junction = heat conducted from junction to B
as we know heat conducted per unit time is given that
Q = thermal conductivity*area*temperature difference/length
390*A*(82 - Tj)/L = 420*A*(Tj - 26)/L
390*(82 - Tj) = 420*(Tj - 26)
390*82 - 390*Tj = 420*Tj - 420*26
Tj*(420+390) = 390*82 + 420*26 = 42900
Tj = 42900/(420+390)
= 52.9629 degree celcius
(b)
using the same method as above and just interchanging the thermal conductivity values:
420*(82 - Tj) = 390*(Tj - 26)
420*82 - 420*Tj = 390*Tj - 390*26
420*82 + 390*26 = Tj*(420 + 390)
Tj = 55.0370 degree celcius
Question (2) Ans.
The rate of change of heat energy can expressed as
Q/t = K*A*(T2 - T1)/L
Where
k = thermal conductivity
A = cross scetional area
T1 = initial temperature
T2 = final temperature
L = thickness of substance
For wool carpet the rate of heat transfer is
Q/t = K*A*(T2 - T1)/L
= (0.04 j/s.m)*(84*10^-4 m^2)*(33 - 10)/0.015 m
= 0.515 W
For ceramic tile the rate of heat transfer is
Q/t = K*A*(T2 - T1)/L
= (0.84 j/s.m)*(84*10^-4 m^2)*(33 - 10)/0.015 m
= 10.7 W