A student heats 10.8 g of lead to a temperature of 90.5 C. He places the hot lead into a container of water, initially at a temperature of 24.6 C; the final temperature of the water and lead is 25.2 C. What quantity of heat was lost by the lead?
heat lost mass Pb x specific heat lead x delta T.
ok i got 0.882, is that correct? How do I calculate the mass of water in the container?
I figured mass of water was next. It's better if you type the whole problem instead of doing it piece meal.
heat lost by Pb + heat gained by water = 0
(mass Pb sp.h Pb x delta T) + (mass water x specific heat water x delta T) = 0.
The first term is the hat lost by lead and the second term is the heat gained by water. The only unknown is the mass of water.
To find the quantity of heat lost by the lead, we can use the formula:
Q = m * c * ΔT
Where:
Q is the quantity of heat lost
m is the mass of the lead
c is the specific heat capacity of lead
ΔT is the change in temperature
First, let's find the change in temperature (ΔT) of the lead:
ΔT = final temperature - initial temperature
ΔT = 25.2 C - 90.5 C
ΔT = -65.3 C
Now, let's calculate the quantity of heat lost by the lead:
Q = m * c * ΔT
The specific heat capacity of lead (c) is 0.128 J/g°C. The mass of the lead (m) is given as 10.8 g.
Q = 10.8 g * 0.128 J/g°C * -65.3 C
Q ≈ -89.13 J
The negative sign indicates that heat was lost by the lead. So, the quantity of heat lost by the lead is approximately 89.13 J.