# 4638

xample question and calculation: You decided to make hot water for your favorite flavored tea drink. The water, after boiling, was at about 98 ?C when you poured it over the tea bags. You now have 100 mL hot tea at 98 ?C and want to dilute the tea and bring it to room temperature at about 25 ?C. To make the tea to the correct temperature, you will add a portion of cold water at a temperature of 5 ?C. How much cold water should you add to the hot tea water? (Hint: the density of tea and water is 1 g/mL and the specific heat of tea and water is 4.184 J/g??C). First convert the volume to mass using density: 100mL?1g1mL=100g100mL?1g1mL=100g Next, use the following equation: q=m×C×?Tq=m×C×?T. This problem requires an endothermic and an exothermic reaction; therefore the equation is modified: q=?qq=?q. The cold water absorbs heat and is the endothermic reaction; therefore, the cold water portion is the qq. The tea temperature being brought down is releasing heat and therefore is the ?q?q side of the equation. Now we can substitute the m×C×?Tm×C×?T for each qq, making sure to keep the signs of each qq. (m×C×?T)=?(m×C×?T)(m×C×?T)=?(m×C×?T) (m×C×(Tf?Ti))=?(m×C×(Tf?Ti))(m×C×(Tf?Ti))=?(m×C×(Tf?Ti)) (m×4.184Jg??×(25??5?))=?(100g×4.184Jg??×(25??98?))(m×4.184Jg??×(25??5?))=?(100g×4.184Jg??×(25??98?)) (m×4.184Jg??×(20?))=?(100g×4.184Jg??×(?73?))(m×4.184Jg??×(20?))=?(100g×4.184Jg??×(?73?)) (m×83.68Jg)=?(?30,543.2J)(m×83.68Jg)=?(?30,543.2J) (m×83.68Jg)=30,543.2J(m×83.68Jg)=30,543.2J m=30,543.2J83.68Jg=365gm=30,543.2J83.68Jg=365g Convert the grams into mL with the density: 365g×1mL1g=365m