What is the specific heat of a metal if addition of 80.0 g of the metal at 24.0

Question

What is the specific heat of a metal if addition of 80.0 g of the metal at 24.0 degrees Celsius to 170.0 g of Cu (s= .38 J/g-c) at 266.4 degrees Celsius produces a mixture that reaches thermal equilibrium at 203.4 degrees Celsius?

s =               J/g-c

The heat capacity of bomb calorimeters are determined by running a known, standard reaction. The combustion of 2.450 mmol of x caused the temperature of a bomb calorimeter to rise from 23.400 degrees Celsius to 26.600 degrees Celsius. For the case where T= 25.000 degrees Celsius, delta H = -2918 kj/mol, and delta ngas= -2 mol gas/ mol x for the combustion reaction, determine the heat capacity of the calorimeter.

C =            kj/c

Explanation / Answer

Heat lost from copper

= mc(deltaT) = (170)(0.38)(266.4-203.4) = 1485.8 J

deltaH = mc(deltaT)

Solve for c (which you wrote as s) using this heat

1485.8 = (80)(c)(203.4-24)

c = 0.1035..... = 0.10

Same as second part of first question. solve for c

deltaH = mc(deltaT)

(2918kJ/mol)(0.00245mol) = C(26.6-23.4)

C = 2234kJ/C

The reason that there is no m in this equation is that it is incorporated into C

Normally c would have unit of J/gC or something like that

but since mass of calorimeter is constant, m and c can be incorporated together for heat capacity??

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.