Question

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

Answer 1

Answer:

$T_{EQ}=42.1\°C$

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:

$Q_{brass}+Q_{water}=0$

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

$m_{brass}C_{brass}(T_{EQ}-T_{brass})+m_{water}C_{water}(T_{EQ}-T_{water})=0$

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

$m_{brass}C_{brass}T_{EQ}+m_{water}C_{water}T_{EQ}=m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}\\\\T_{EQ}(m_{brass}C_{brass}+m_{water}C_{water})=m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}\\\\T_{EQ}=}\frac{m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}}{m_{brass}C_{brass}+m_{water}C_{water}}$

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

$T_{EQ}=\frac{130g*4.184\frac{J}{g*\°C} *35.0\°C+130g*0.380\frac{J}{g*\°C}*120.0\°C}{130g*4.184\frac{J}{g*\°C} +130g*0.380\frac{J}{g*\°C}}\\\\T_{EQ}=42.1\°C$

Best regards!