Question

A square brass plate with sides Lo = 15.0 cm at 67 °F is placed in a hot water bath, maintained at 85 'F. Follow the steps below to derive a formula for fractional increase in area AA/A, and use the formula to calculate AA=A-A=L? - Lo and % change in area. [L=L(1 + .AT); a = 2.0 x 10K for brass. ] (5 pts) Step 1: Obtain an expression for A involving A, a, and AT: A=L= Step 2: Drop any term involving a which will be negligible compared to others. Explain why. Rearrange to find equations for AA and AA/A. (5 pts) AA = AA/A. = Step 3: Write the formula for converting AT from Fahrenheit (ATF) to Celsius (ATC). How is it different from the formula: Te -(5/9) (Tp-32)? Find AT for our problem in C and K: (5 pts.) Formula: (ATC) AT Step 4: Calculate the increase in area, AA, of the plate and the fractional increase AA/Ao as a %. (5 pts.)

Answer 1

step 1)

A = L = (L.(1 + QAT))2 = 1 (1+QAT) = A (1+QAT)

step 2)

Α = Αρ(1 + 2α ΔΤ +a2ΔΤ2)

a? <<1
so the term
APAT
can be neglected.

A = 4(1+20 AT)

ΔΑ = A - Αρ = Αρ (1 + 2αΔΤ) – Αρ = Αρ(2αΔΤ)

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step 3)

----(1)

Tc = -(Tp – 32)

for a small change in temperature, ------(2)

Tc + ATC = -1; + ATF – 32)

(2) -(1) gives

ATC= AT:

Change in temperature in degree Celsius is

ATC= AT:

For the given temperatures,  

ATF = 85 – 67 = 18ºF

ATC = -ATE = 2 * 18 = 10°C

ATC = 10K

step 4)

Lo = 15.0 cm = 0.15 m

For brass

a = 2.0 * 10-5 K-1

AA = L(20AT) = (0.15)-(2 * 2.0 * 10-5 * 10) = 9 * 10-m?

* 100 = (20 AT) * 100 = (2 * 2.0 * 10-5 * 100 * 100 = 0.04% Ар