A density of 7.90 g/cm^3 A specific heat of 0.450 J/g-K A coefficient of linear

Question

A density of 7.90 g/cm^3 A specific heat of 0.450 J/g-K A coefficient of linear expansion of 13.5 times 10-6/C degree It's consider his left femurs to be a cylinder with the following properties when Wolverine is at rest: - A radius of 1.25 cm -A length of 50.5 cm -A mass of 1980 g During a certain fight, Wolverine's core body temperature increases from 38.0 degree C to 39.0 degree C. What is the change in length (Delta L) of his demure given the temperature change his body experiences. Delta L = 6.82 times 10^-4 How much thermal energy (Delta E_Th) does the femora absorb during the temperature change? Delta E_Th = 891

Explanation / Answer

dL = L*alpha*dT

dT = 1 C

alpha = 13.5*10^-6 /C

L = 0.505 m

dL = 0.505*13.5*10^-6*1

dL = 6.817*10^-6 m = 6.82*10^-4 cm

B.

E = m*Cp*dT

E = 1980 g*0.45 J//g-C*1 C = 891 J

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