GO Figure 18-46 shows the cross section of a wall made of three layers. The layer thicknesses are L 1 , L 2 = 0.700 L 1 , and L 3 = 0.350 L 1 . The thermal conductivities are k 1 , k 2 = 0.900 k 1 , and k 3 = 0.800 k 1 . The temperatures at the left side and right side of the wall are T H = 30.0°C and T C = −15.0°C, respectively. Thermal conduction is steady. (a) What is the temperature difference Δ T 2 across layer 2 (between the left and right sides of the layer)? If k 2 were, instead, equal to 1.1 k 1 , (b) would the rate at which energy is conducted through the wall be greater than, less than, or the same as previously, and (c) what would be the value of Δ T 2 ? Figure 18-56 Problem 60.
GO Figure 18-46 shows the cross section of a wall made of three layers. The layer thicknesses are L 1 , L 2 = 0.700 L 1 , and L 3 = 0.350 L 1 . The thermal conductivities are k 1 , k 2 = 0.900 k 1 , and k 3 = 0.800 k 1 . The temperatures at the left side and right side of the wall are T H = 30.0°C and T C = −15.0°C, respectively. Thermal conduction is steady. (a) What is the temperature difference Δ T 2 across layer 2 (between the left and right sides of the layer)? If k 2 were, instead, equal to 1.1 k 1 , (b) would the rate at which energy is conducted through the wall be greater than, less than, or the same as previously, and (c) what would be the value of Δ T 2 ? Figure 18-56 Problem 60.
GO Figure 18-46 shows the cross section of a wall made of three layers. The layer thicknesses are L1, L2 = 0.700L1, and L3 = 0.350L1. The thermal conductivities are k1, k2 = 0.900 k1, and k3 = 0.800 k1. The temperatures at the left side and right side of the wall are TH = 30.0°C and TC = −15.0°C, respectively. Thermal conduction is steady. (a) What is the temperature difference ΔT2 across layer 2 (between the left and right sides of the layer)? If k2 were, instead, equal to 1.1k1, (b) would the rate at which energy is conducted through the wall be greater than, less than, or the same as previously, and (c) what would be the value of ΔT2?
3. The entropy of an ideal gas is S = Nkg In V. Entropy is a state function rather than a path
function, and in this problem, you will show an example of the entropy change for an ideal gas
being the same when you go between the same two states by two different pathways.
A. Express ASV = S2 (V2) - S₁(V1), the change in entropy upon changing the volume from V₁to
V2, at fixed particle number N and energy, U.
B. Express ASN = S₂(N₂) - S₁ (N₁), the change in entropy upon changing the particle number
from N₁ to N2, at fixed volume V and energy U.
C. Write an expression for the entropy change, AS, for a two-step process (V₁, N₁) →
(V2, N₁) → (V2, N₂) in which the volume changes first at fixed particle number, then the particle
number changes at fixed volume. Again, assume energy is constant.
Please don't use Chatgpt will upvote and give handwritten solution
6. We used the constant volume heat capacity, Cv, when we talked about
thermodynamic cycles. It acts as a proportionality constant between energy and
temperature: dU = C₁dT.
You can also define a heat capacity for constant pressure processes, Cp. You can think
of enthalpy playing a similar role to energy, but for constant pressure processes
δαρ
C = (37) - Sup
Ср
ат
P
=
ат
Starting from the definition of enthalpy, H = U + PV, find the relationship between Cy
and Cp for an ideal gas.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.