1. The problem statement, all variables and given/known data your 200 gram cup of tea is boiling hot. About how much ice should you add to bring it down to 65C? Ice is initially at -15C. The specific heat capacity of ice id .5cal/gC 3. The attempt at a solution So the ice we add needs to absorb 7000 calories of heat. Heat capacity is Q/T heat divided by change in temperature . And specific heat capacity is is heat capacity divided by mass. So to heat the ice to 0 Celsius we have 15C(.5cal)/gC(x) + 65(x) + 80 cal/g(x) =7000 calories. cleaning it up (15)(.5)x+65x+80x=7000 in units of energy the 80 comes from the latent heat for melting ice which is 80cal/gram and x is the unknown mass of ice. and i get that x=45.9 grams of ice.