Will someone show me the formula or help me through working this question? Thanks!
A 2.85g lead weight, initially at 10.3 deg C, is submerged in 7.55g of water at 52.3 deg C in an insulated container. what is the final temperature of both substances at thermal equilibrium?
formula? The concept is the closed container gains no heat from outside, so the sum of the heats gained is zero (one gains, one loses).
Heatgaindlead+heatgainedwater=0
2.85g*clead*(Tf-10.3C)+7.55g*cwater*(Tf-52.3)=0
look up the two specific heats (watch units, your mass is in grams). Solve for Tfinal.
Alright thanks! Using what you gave me I got T= 53 degrees, hope that's right!
I don't think so....the water was initially hotter, and putting cold lead in made it hotter? I don't think so. Recheck.
To find the final temperature of both substances at thermal equilibrium, we can use the principle of heat transfer. The heat gained by one substance must be equal to the heat lost by the other substance.
Let's start by calculating the heat gained or lost by the lead weight using the formula:
Q = mcΔT
where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For the lead weight:
m = 2.85g
c = specific heat capacity of lead
ΔT = final temperature - initial temperature
Since the specific heat capacity of lead is 0.128 J/g°C, we can calculate the heat gained or lost by the lead weight.
Q_lead = (2.85g) * (0.128 J/g°C) * (final temperature - 10.3°C)
Next, let's calculate the heat gained or lost by the water:
m = 7.55g
c = specific heat capacity of water
ΔT = final temperature - initial temperature
Since the specific heat capacity of water is 4.18 J/g°C, we can calculate the heat gained or lost by the water.
Q_water = (7.55g) * (4.18 J/g°C) * (final temperature - 52.3°C)
According to the principle of heat transfer, Q_lead = -Q_water (since heat is transferred from the lead weight to the water).
Setting the two equations equal to each other:
(2.85g) * (0.128 J/g°C) * (final temperature - 10.3°C) = -(7.55g) * (4.18 J/g°C) * (final temperature - 52.3°C)
Now, we can solve for the final temperature of both substances at thermal equilibrium. Rearranging the equation:
(2.85g) * (0.128 J/g°C) * final temperature - (2.85g) * (0.128 J/g°C) * 10.3°C = -(7.55g) * (4.18 J/g°C) * final temperature + (7.55g) * (4.18 J/g°C) * 52.3°C
Simplifying the equation:
(2.85g) * (0.128 J/g°C) * final temperature + (7.55g) * (4.18 J/g°C) * final temperature = (7.55g) * (4.18 J/g°C) * 52.3°C - (2.85g) * (0.128 J/g°C) * 10.3°C
Solving for final temperature:
10.3 * (2.85 * 0.128 + 7.55 * 4.18) = (7.55 * 4.18 - 2.85 * 0.128) * final temperature
final temperature = [10.3 * (2.85 * 0.128 + 7.55 * 4.18)] / [(7.55 * 4.18 - 2.85 * 0.128)]
After substituting the values and performing the calculation, you will find the final temperature of both substances at thermal equilibrium.