Question

three lead fishing weights, each with a mass of 1.00x10^2 g and at a temperature of 100.0oC, are placed in 2.00x10^2 g of water at 35.0oC. The final temperature of the mixture is 45.0oC. What is the specific heat of the lead in the weights?

Answer 1

The specific heat of the lead in the weights is 0.49 J/g °C.

Step-by-step explanation:

To find the specific heat of the lead in the weights, we can use the equation:

Q = mcΔT

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, let's find the heat transferred to the water:

Qwater = mwater * cwater * ΔTwater

Qwater = (200.0g) * (4.18 J/g °C) * (45.0 °C - 35.0 °C)

Qwater = 8360 J

Since the water and the lead reach thermal equilibrium, the heat transferred to the water is equal to the heat transferred from the lead:

Qwater = Qlead

Qlead = mlead * clead * ΔTlead

We know the mass of the lead (3 * 1.00x10^2 g = 300.0g), the change in temperature (45.0 °C - 100.0 °C = -55.0 °C), and the heat transferred to the water (8360 J).

Now we can solve for the specific heat of the lead:

clead = Qlead / (mlead * ΔTlead)

clead = 8360 J / (300.0g * -55.0 °C)

clead = 0.49 J/g °C

Answer 2

The lead and the water reach thermal equilibrium when the net flow of heat from the lead to the water becomes zero, so when the heat released by the lead
`Q_L` is equal to the heat absorbed by the water
`Q_w`:

`-Q_L = Q_w` (1)
where the negative sign means the heat of the lead is released, not absorbed.

The equation that gives the heat absorbed/released is

`Q=m C \Delta T` (2)
where
m is the mass of the substance
C is its specific heat capacity

`\Delta T` is the variation of temperature.

If we rewrite (1) by using (2), we get

`-m_L C_L \Delta T_L = m_w C_w \Delta T_w` (3)
where the label L refers to the lead and the label W refers to the water.

The total mass of the lead is:

`m_L = 3 \cdot 1.00 \cdot 10^2 g =300 g`
while its change of temperature is

`\Delta T_L = T_f - T_i = 45.0^(\circ) C - 100^(\circ)C = -55^(\circ) C`

The mass of the water is

`m=2.00 \cdot 10^2 g = 200 g`
its specific heat capacity is

`C_w = 4.18 J/(g ^(\circ)C)`
while its change of temperature is

`\Delta T_w = T_f - T_i = 45.0^(\circ) C- 35.0^(\circ) C= 10^(\circ)C`

If we put all this data inside (3), we can calculate the specific heat capacity of lead:

`C_L = - (m_w C_w \Delta T_w)/(m_L C_L)=- ((200 g)(4.18 J/g^(\circ)C)(10^(\circ)C))/((300 g)(-55^(\circ)C))= 0.50 J/(g^(\circ)C)`