In: Chemistry
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 114.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.
1. q1 = mCt
= 13*2.09*(10) = 271.7J
q2 = fustion*m = 333.55*13 = 4336.15J
q3 = mct
= 13*4.184*(100) = 5439.2J
q4 = heat of vaporisation * m
= 2260*13 = 29380J
q5 = mct
= 13*2.01*(114-100)
= 365.82J
total heat = q1 + q2+ q3+q4+q5 = 271.7 + 4336.15+5439.2 +29380 + 365.82 = 39792.87J = 39.793KJ