Question

4. Specific Heat. A 900-g brass at 100 o C is dropped into a 2.27-kg water at 1.60oC. When mixed, the equilibrium temperature is 5.10 oC. Find the specific heat of brass. (specific heat of water = 1.00 cal/g∙C0)

5. First Law of Thermodynamics. You do 10,000 J of work on a system of 3.00 kg water by stirring it with a paddle wheel. During this time, 10,000 calories of heat is added. What is change in the internal energy of the system? (1 calorie = 4.18 Joules) 

Answer 1

4)

$$m_{brass}c_{brass}(t_{brass}-t_f)=m_{w}c_{w}(t_{f}-t_{w}),$$ $$c_{brass}=\dfrac{m_{w}c_{w}(t_{f}-t_{w})}{m_{brass}(t_{brass}-t_f)},$$ c_{brass}=\dfrac{2270\ g\cdot1.0\ \dfrac{cal}{g\cdot\!^{\circ}C}\cdot(5.1^{\circ}C-1.6^{\circ}C)}{900\ g\cdot(100^{\circ}C-5.1^{\circ}C)}=0.093\ \dfrac{cal}{g\cdot\!^{\circ}C}.

5) By the First Law of Thermodynamics we have:

$$\Delta U=Q+W,$$ $$\Delta U=10000\ cal\cdot\dfrac{4.18\ J}{1\ cal}+10000\ J=51800\ J.$$