Answer :
Qt = 4.719kJ heat, is required to warm 10.0 g of ice, initially at -10.0 degrees Celsius, to steam at 110 degrees celsius. The heat capacity of ice is, c = 2.09 J/g C.
The mass of ice is, m = 10.0 g.
The initial temperature is, T = - 10 degree Celsius.
The final temperature is, T' = 110.0 degree Celsius.
The heat capacity of ice is, c = 2.09 J/g C.
The heat capacity of steam is, c' = 2.01 J/g C.
When ice absorbs heat, the conversion takes place from ice to water (liquid) and from the liquid to steam at 110 degree Celsius. So,
The amount of heat absorbed is given as,
Q = m[c(T' - T) + c'T']
Q = 10[2.09(110-(-10)+(2.01*110)]
Q = 4719J
Convert in kJ as,
Q = 4719/1000
Q = 4.719 kJ
Thus, we can conclude that the amount of heat required to warm the ice is 4.719 kJ.
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The complete question is -
How much heat (in kJ) is required to warm 10.0 g of ice, initially at −10.0 °C, to steam at 110.0 °C? The heat capacity of ice is 2.09 J/g · °C, and that of steam is 2.01 J/g °C.