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aal0315
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Homework Statement
A block of ice at 0C is added to a 150g aluminum calorimeter cup which holds 210g of water at 12C. If all but 2.0 g of ice melt, what was the original mass of the block of ice
Homework Equations
heat gained by ice = heat lost by water + heat lost by calorimeter
The Attempt at a Solution
so i figure that its:
(m(i)+2.0x10^-3kg)*c(i)*deltaT(i) = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a), where m is mass, c is the specific heat and T is temperature
but what i don't get is isn't the deltaT(i) going to be zero since the ice starts off zero and the water/ice mixture temperature zero? so then this equation doesn't work because solving for m(i), you can't divide a number by zero?