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A 0.095-kg aluminium sphere is dropped from the roof of a 55-m-high building. The specific heat of aluminium is 900 J/kg⋅C∘ .
If 65 % of the thermal energy produced when it hits the ground is absorbed by the sphere, what is its temperature increase?

Respuesta :

Answer:

Increase in temperature will be [tex]0.389^{\circ}C[/tex]

Explanation:

We have given mass of the aluminium m = 0.095 kg

Height h = 55 m

Specific heat of aluminium c = 900 J/kg°C

We know that potential energy is given as

[tex]PE=mgh=0.095\times 9.8\times 55=51.205[/tex]

Now 65 % of potential energy [tex]=\frac{51.205\times 65}{100}=33.28[/tex]

Now this energy is used to increase the temperature

So [tex]mc\Delta T=33.28[/tex]

[tex]0.095\times 900\times \Delta T=33.28[/tex]

[tex]0.095\times 900\times \Delta T=33=0.389^{\circ}C[/tex]

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