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Calorimetry and Hess' Law

Detailed work would be very helpful; the math involved in this gets me confused, but I'm not sure where I have been going wrong. If work is handwritten, please do not write in cursive for I have problems reading it.

1. The mass of a substance is 200gm and its specific heat is 0.09. How much heat is required to raise the temperature of the substance from 20°C to 90°C?

2. The heat capacities of mercury and glass are the same. The density of mercury is 13.6gm/cc and the density of glass is 2.5gm/cc. If the specific heat of mercury is 0.03, what will be the specific heat of glass?

3. A body of mass 100gm is heated to 122° C and then quickly dropped into a copper calorimeter of mass 50gm. The calorimeter contains 300gm of water at 28° C. The final temperature of the system becomes 30° C. What is the specific heat of the body if the specific of copper is 0.09?

4. Use the following information to determine the enthalpy (?H°) of the reaction for:
NO2 (g) + (7/2) H2 (g) --> 2H2O (l) + NH3 (g)
Using the following two equations:
2NH3 (g) --> N2 (g) + 3H2 (g) ?H° = +92 kJ
(1/2) N2 (g) + 2H2O (l) --> NO2 (g) + 2H2 (g) ?H° = +170kJ

5. The compound carbon suboxide, C3O2, is a gas at room temperature. Use the data supplied the calculate the heat of formation of carbon suboxide.
2CO (g) + C (s) --> C3O2 (g) ?H° = +127.3 kJ
CO(g) ?H° = -110.5 kJ

Solution Preview

1. The mass of a substance is 200gm and its specific heat is 0.09. How much heat is required to raise the temperature of the substance from 20° C to 90° C?
If the unit for specific heat is cal/(gm0C), unit for the mass is gm and unit for ∆T is 0C.
Therefore, Q=mS∆T=200*0.09*(90-20)=1260 cal (units of gm and 0C have been canceled out).

2. The heat capacities of mercury and glass are the same. The density of mercury is 13.6gm/cc and the density of glass is 2.5gm/cc. If the specific heat of mercury is 0.03, what will be the specific heat of glass?
Assume V is the volume for mercury and glass, s is the specific heat for glass.
Then the ...

Solution Summary

The expert examines calorimetry and Hess' law.

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