A- a small volume of cold water is used to cool off a cup of coffee that is init

Question

A- a small volume of cold water is used to cool off a cup of coffee that is initially too hot to drink. How many mL of water at 4.0oC must be mixed with 485 mL of hot coffee (95.0oC) so that the resulting combination has a temperature of 70.0oC? Assume that the coffee and water have the same density and specific heat.   

B- A 5.00 cm3 sample of aluminum at room temperature (25.00oC) was dropped into a 1 L beaker of boiling water (100.00oC) and allowed to come to equilibrium. How much heat was absorbed by the aluminum cube? The specific heat capacity of aluminum is 0.902 J/g·K and its density is 2.70 g/cm3.

C- When 50.00 g of 0.200 M NaCl (aq) at 24.1oC is added to 100.0 g of 0.100 M AgNO3 (aq) at 24.1oC in a calorimeter, the temperature increases to 25.2oC as AgCl (s) forms. Assuming the specific heat of the solution and products is 4.20 J/goC, calculate the number of joules of heat produced in the reaction.

Explanation / Answer

A- a small volume of cold water is used to cool off a cup of coffee that is initially too hot to drink. How many mL of water at 4.0oC must be mixed with 485 mL of hot coffee (95.0oC) so that the resulting combination has a temperature of 70.0oC? Assume that the coffee and water have the same density and specific heat.   

T = 4

V = 485

T = 95

Tf = 70

Qwin = -Qlost

Qwin = mc*Cp*(Tf-Tc)

Qlost = mh*Cp*(Tf-Th)

mc*Cp*(Tf-Tc) = - mh*Cp*(Tf-Th)

mc*(70-4) = -485*(70-95)

mc = 485(25/66) = 183.7 grams of 183.7 ml

V = 183.7 ml

B- A 5.00 cm3 sample of aluminum at room temperature (25.00oC) was dropped into a 1 L beaker of boiling water (100.00oC) and allowed to come to equilibrium. How much heat was absorbed by the aluminum cube? The specific heat capacity of aluminum is 0.902 J/g·K and its density is 2.70 g/cm3.

V = 5 cm3 of Aluminium

T = 25

V = 1L

T = 100

Qheat = ?

Cp Al = 0.902

D = 2.7 g/cm3

Qal = mal*Cpal*(Tf-Ti)

Assume final temperature to be 100, Ti = 25

m = D*V = 2.70*5 = 13.5 g of Al

Q = 13.5*0.902*(100-25) = 913.3 J

C- When 50.00 g of 0.200 M NaCl (aq) at 24.1oC is added to 100.0 g of 0.100 M AgNO3 (aq) at 24.1oC in a calorimeter, the temperature increases to 25.2oC as AgCl (s) forms. Assuming the specific heat of the solution and products is 4.20 J/goC, calculate the number of joules of heat produced in the reaction.

m = 50 g of solution

M = 0.2 NaCl

T = 24.1

m = 100 g

M = 0.1 AgNO3

T = 24.1

Tf = 25.2

Cp = 4.20

find heat of reaction per reaction

Q = m*Cp*(Tf-Ti)

Q = (50+100)*(4.20)(25.2-24.1) = 693 J

find moles of Ag and moles of Cl

assume D = 1g/ml so

50 g --> 50 ml of solution and 100 g are 100 ml of solution

mol of Ag = M*V = 100*0.1 = 10 mmol of Ag

mol of Cl = M*V = 0.2*50 = 10 mmol of Cl

therefore basis is 10 mmol

E = Q/n = 693 J/(10*10^-3 mol) = 69600 J/mol

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