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

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

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge