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×ΔTq=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×ΔTm×C×∆T for each qq, making sure to keep the signs of each qq.
(m×C×ΔT)=−(m×C×ΔT)(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
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×ΔTq=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×ΔTm×C×∆T for each qq, making sure to keep the signs of each qq.
(m×C×ΔT)=−(m×C×ΔT)(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
