lab chemistry

xample inquiry and calculation:

You unwavering to compel hot instil for your jewel flavored tea imbibe. The instil, forthcoming paroxysm, was at environing 98 ˚C when you poured it balance the tea bags. You now entertain 100 mL hot tea at 98 ˚C and deficiency to enervate the tea and convey it to extent air at environing 25 ˚C. To compel the tea to the amend air, you conciliate add a concern of indifferent instil at a air of 5 ˚C. How plenteous indifferent instil should you add to the hot tea instil? (Hint: the dullness of tea and instil is 1 g/mL and the particular excitement of tea and instil is 4.184 J/g∙˚C).

First transmute the quantity to heap using dullness:

100mL⋅1g1mL=100g100mL⋅1g1mL=100g

Next, use the forthcoming equation: q=m×C×ΔTq=m×C×∆T. This drift requires an endothermic and an exothermic reaction; accordingly the equation is modified: q=−qq=−q. The indifferent instil absorbs excitement and is the endothermic reaction; accordingly, the indifferent instil concern is the qq. The tea air nature brought down is releasing excitement and accordingly is the −q−q verge of the equation. Now we can supply the m×C×ΔTm×C×∆T for each qq, making indisputable to celebrate 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 after a while the dullness: 365g×1mL1g=365m
 

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