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Hunter-Best [27]
2 years ago
14

In the experiment to measure specific heat of a metal, you get the following data. What is the specific heat of the metal in J/k

g.C?
(answer within +-3% will considered correct) MASSES: water= 259 g, metal=159 g, calorimeter=97 g INITIAL TEMPERATURES: water+calorimeter=21ºC, metal=98ºC Final equilibrium temp=31ºC Calorimeter is made of aluminum.
Physics
1 answer:
crimeas [40]2 years ago
7 0

Answer:

c = 1,100 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*97g*10ºC + 4.186 J/gºC*259g*10ºC = 11,715 J(1)

Ql = cₓ*159g*67ºC   (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = 11,715 J / 159g*67ºC = 1.1 J/gºC (3)

Expressing (3) in J/kgºC:

1.1 J/gºC * (1,000g/1 kg) = 1,100 J/kgºC  

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Answer:

a) Spring Constant K=18.946\ \rm N/m

b) Amplitude of the oscillationA=0.3\ \rm m

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Explanation:

Given:

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a)

When Finally the mass comes to rest after 30 cm of stretching in spring then work done by all the forces is equal to the change in kinetic Energy of the of the mass

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\nu=\sqrt{\dfrac{K}{m}}\\\nu=\dfrac{1}{2\pi}\sqrt{\dfrac{18.946}{0.29}}\\\nu=1.287\ \rm rev/s

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Explanation:

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