The mass of ice required at 0oc that 10 grams of 100oc steam will melt completely is 80.07 g.
The amount of energy that must be supplied to a solid substance (typically in the form of heat) in order to cause a change in its physical state and convert it to a liquid is known as latent heat of fusion (when the pressure of the environment is kept constant). The latent heat of fusion of one kilogramme of water, for example, is 333.55 kilojoules, which is the amount of heat energy that must be supplied to convert one kilogramme of ice without changing the temperature of the environment (which is kept at zero degrees Celsius).
Given,
mass of steam, [tex]m_s[/tex] = 10g = 0.01 kg
Latent heat of fusion of ice ,[tex]L_f= 334kj/kg[/tex]
Latent heat of vaporization of water, [tex]L_v=2256kj/kg[/tex]
change of temperature, [tex]\delta T=100-0=100\textdegree\ C[/tex]
The amount of heat cooling can be determined by.
[tex]m_{ice}L_f=m_sL_v+m_sS \delta T[/tex]
Here, S= specific heat constant i.e [tex]4.184 kj/kg[/tex]
[tex]m_{ice}L_f=(0.01*2256)+(0.01*4.184*100)\\\\m_{ice}*334=22.56+4.184\\\\m_{ice}*334=26.744\\\\m_{ice}=\frac{26.744}{334}=0.0807kg=80.07 g[/tex]
Thus, the mass of quantity of ice is 80.07 g.
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