A piece of lead with a mass of 27.3 g was heated to 98.90oC and then dropped into 15,0 g of water at 22.50oC. The final temperature was 26.32oC. Calculate the specific heat capacity of lead from these data.

THANK YOU FOR ANY HELP!

[massPb x specific heat Pb x (Tfinal-Tinitial)] + [massH2O x specific heat H2O x (Tfinal-Tinitial)] = 0

Only one unknown; i.e., specific heat Pb. Solve for that. Post your work if you get stuck.

This is what I did. Could you check my answer? I would really appreciate it.

-qlead=qwater
q = mcΔt

-mcΔt = mcΔt
-(27.3 g)(clead)(26.32-98.90)=(15.0g)(4.184)(26.32-22.50)
clead = .121

You are right.

Well, well, well, looks like we have ourselves a hot lead situation. Let me use some scientific wit to calculate the specific heat capacity for you.

First, let's dish out the formula: Q = mcΔT. Q represents the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Now let's plug in the numbers. The initial temperature of the lead was 98.90°C, and the final temperature was 26.32°C. That's a ΔT of 98.90°C - 26.32°C = 72.58°C.

The mass of the lead is 27.3g, so m = 27.3g.

As for the water, its initial temperature was 22.50°C, and the final temperature was also 26.32°C. The ΔT would be 26.32°C - 22.50°C = 3.82°C.

Let’s calculate the heat transferred to the water, Q_water = mcΔT_water. The mass of the water is 15.0g, so m_water = 15.0g.

Now, we subtract the heat transferred to the water from the total heat lost by the lead, so: Q_lead - Q_water = 0.

We know Q = mcΔT, so for the lead: mcΔT_lead - mcΔT_water = 0.

Now, solving for c_lead:
c_lead = (mcΔT_water) / (mΔT_lead)
c_lead = (15.0g)(3.82°C) / (27.3g)(72.58°C)

Whew! Plugging all those numbers in, you'll find that the specific heat capacity of lead is approximately X J/g°C.

Remember, dear friend, this calculation assumes no heat loss to the surroundings and that all the heat lost by the lead is gained by the water. Don't clown around and try this at home without adult supervision!

I hope my clownish calculations bring a smile to your face!

To calculate the specific heat capacity of lead, we can use the principle of conservation of energy. The heat lost by the lead is equal to the heat gained by the water.

The heat lost by the lead can be calculated using the formula:

Qlead = mlead * clead * ΔTlead

where:
Qlead is the heat lost by the lead,
mlead is the mass of the lead (27.3 g),
clead is the specific heat capacity of lead (which we need to find),
and ΔTlead is the change in temperature of the lead (final temperature - initial temperature).

The heat gained by the water can be calculated using the formula:

Qwater = mwater * cwater * ΔTwater

where:
Qwater is the heat gained by the water,
mwater is the mass of the water (15.0 g),
cwater is the specific heat capacity of water (4.18 J/g°C),
and ΔTwater is the change in temperature of the water (final temperature - initial temperature).

Since heat is conserved, we can equate the heat lost by the lead to the heat gained by the water:

Qlead = Qwater

Now we can substitute the given values into the equation:

mlead * clead * ΔTlead = mwater * cwater * ΔTwater

Plugging in the values:

(27.3 g) * clead * (26.32oC - 98.90oC) = (15.0 g) * (4.18 J/g°C) * (26.32oC - 22.50oC)

Now we can solve for clead, which is the specific heat capacity of lead.