Answered

Join IDNLearn.com and start exploring the answers to your most pressing questions. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.

A 130 g sample of brass at 120.0 degrees Celsius is placed in a calorimeter cup that contains
130 g of water at 35.0 degrees Celsius. Disregard the absorption of heat by the cup and
calculate the final temperature of the brass and water. Specific heat of water = 4.18 J/gC,
specific heat of brass=0.380 J/gC. Attach your complete solution

Sagot :

Sebassandinidn

Answer:

[tex]T_{EQ}=42.1\°C[/tex]

Explanation:

Hello there!

In this case, since we are looking for an equilibrium temperature, we can evidence how the net heat flow is 0 as all the heat released by the hot brass is absorbed by the water:

[tex]Q_{brass}+Q_{water}=0[/tex]

Thus, by writing that equation in terms of mass, specific heat and temperature we obtain:

[tex]m_{brass}C_{brass}(T_{EQ}-T_{brass})+m_{water}C_{water}(T_{EQ}-T_{water})=0[/tex]

Thus, by applying the following algebra, it is possible to arrive to a clean expression to obtain the equilibrium temperature:

[tex]m_{brass}C_{brass}T_{EQ}+m_{water}C_{water}T_{EQ}=m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}\\\\T_{EQ}(m_{brass}C_{brass}+m_{water}C_{water})=m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}\\\\T_{EQ}=}\frac{m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}}{m_{brass}C_{brass}+m_{water}C_{water}}[/tex]

Then, by plugging the masses, specific heats and initial temperatures in, we obtain:

[tex]T_{EQ}=\frac{130g*4.184\frac{J}{g*\°C} *35.0\°C+130g*0.380\frac{J}{g*\°C}*120.0\°C}{130g*4.184\frac{J}{g*\°C} +130g*0.380\frac{J}{g*\°C}}\\\\T_{EQ}=42.1\°C[/tex]

Best regards!

Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.