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3 years ago

How much heat is required to convert 50 g of ice at -5 degrees Celsius to steam?

Physics

1 answer:

12345 [234]3 years ago

3 0

Answer:

15435 J

Explanation:

Latent heat of fusion, Lf = 334 J/g

Specific heat of ice, ci = 2.1 J / g C

Latent heat of vaporisation, Lv = 2230 J/g

Specific heat of water, cw = 4.18 J / g C

mass, m = 50 g, T = - 5 degree C

There are following steps

(i) - 5 degree C ice converts into 0 degree C ice

H1 = m x ci x ΔT = 50 x 2.1 x 5 = 525 J

(ii) 0 degree C ice converts into 0 degree C water

H2 = m x Lf = 5 x 334 = 1670 J

(iii) 0 degree C water converts into 100 degree C water

H3 = m x cw x ΔT = 5 x 4.18 x 100 = 2090 J

(iv) 100 degree C water converts into 100 degree C steam

H4 = m x Lv = 5 x 2230 = 11150 J

Total heat required

H = H1 + H2 + H3 + H4

H = 525 + 1670 + 2090 + 11150 = 15435 J

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Answer:

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Explanation:

We are given that

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Initial momentum of system=0

$m_1v_1=-m_2v_2$

$v_1=-\frac{m_2v_2}{m_1}$

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

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K.E of shell= $\frac{1}{2}(\frac{m^2_2v^2_2}{m_1}+\frac{1}{2}m_2v^2_2)$

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Velocity of shell fired from this loose cannon,v_2= $\sqrt{\frac{2k.E}{m_2(\frac{m_2}{m_1}+1)}$

$v_2=\sqrt{\frac{2\times 1.09\times 10^7}{73.5(\frac{73.5}{5.71\times 10^3}+1)}$

$v_2=541.14m/s$

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