(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

heat gained by ice = heat lost by water + heat lost by calorimeter

3. 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 dont get is isnt the deltaT(i) going to be zero since the ice starts off zero and the water/ice mixture temperature zero? so then this equation doesnt work because solving for m(i), you cant divide a number by zero?

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# Homework Help: Calorimetry with ice and water

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