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Sholpan [36]
3 years ago
11

If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

at capacity of the metal?
Physics
2 answers:
kobusy [5.1K]3 years ago
8 0

sorry i don't know the answer i'm really sorry

forsale [732]3 years ago
6 0
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
where m is the mass of the substance, Cs is its specific heat capacity and \Delta T is the increase of temperature.

If we re-arrange the formula, we get
C_s =  \frac{Q}{m \Delta T}
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
C_s =  \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C
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Answer:

42000N

Explanation:

First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.

1) linear thermal expansion coef brass 19e-6 /K

∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm

Second part involves linear elasticity.

for brass, young's modulus is 15e6 psi or 100 GPa

cross-sectional area of rod is π(0.008)² = 0.0002 m²

F = EA∆L/L

F = (100e9)(0.0002)(0.00387) / (1.85)

F = 42000 or 42 kN

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Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial
Maru [420]

Our data are,

State 1:

P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K

State 2:

P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3

We know as well that 3BTU=3.16kJ/K

To find the mass we apply the ideal gas formula, which is given by

P_1V_1=mRT_1

Re-arrange for m,

m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Replacing,

T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F

For conservative energy we have, (Cv = 0.718)

W = m C_v = 0.718  \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU

3 0
3 years ago
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