Equal 0.400-kg mass of lead and tin at 60 degrees celsius are placed in 1kg of water at 20 degrees celsius.

Question

a) what is the equilibrium temperature of the system?
b)if an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy?
c) how many atoms of tin are in 0.400kg of tin and how many atoms of lead are in 0.400kg of lead?
d) divide the number of tin atoms by the number of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. what conclusion can be drawn?

Answer 1

a) To find the equilibrium temperature of the system, we need to use the conservation of energy principle. The heat gained by water should be equal to the heat lost by the lead and tin:

Q_water = Q_lead + Q_tin

We know that Q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature.

For water, m_water = 1 kg, c_water = 4186 J/(kg·K), and ΔT_water = T_eq - 20 °C.

For lead, m_lead = 0.400 kg, c_lead = 128 J/(kg·K), and ΔT_lead = 60 °C - T_eq.

For tin, m_tin = 0.400 kg, c_tin = 227 J/(kg·K), and ΔT_tin = 60 °C - T_eq.

Now plug in the values and solve for T_eq:

(1 kg)(4186 J/(kg·K))(T_eq - 20 °C) = (0.400 kg)(128 J/(kg·K))(60 °C - T_eq) + (0.400 kg)(227 J/(kg·K))(60 °C - T_eq)

T_eq ≈ 30.64 °C

b) If an alloy is half lead and half tin by mass, the specific heat can be found using the weighted average of the specific heats:

c_alloy = (0.5)(c_lead) + (0.5)(c_tin) ≈ (0.5)(128 J/(kg·K)) + (0.5)(227 J/(kg·K)) ≈ 177.5 J/(kg·K)

c) To find the number of atoms in a mass, we can use the formula:

Number of atoms = (mass × Avogadro's number) / molar mass

For tin:

Number of tin atoms = (0.400 kg × 6.022 × 10^23 atoms/mol) / (118.71 g/mol × 1 kg/1000 g) ≈ 2.03 × 10^24 atoms

For lead:

Number of lead atoms = (0.400 kg × 6.022 × 10^23 atoms/mol) / (207.2 g/mol × 1 kg/1000 g) ≈ 1.16 × 10^24 atoms

d) Divide the number of tin atoms by the number of lead atoms:

(2.03 × 10^24 atoms) / (1.16 × 10^24 atoms) ≈ 1.75

Now divide the specific heat of tin by the specific heat of lead:

(227 J/(kg·K)) / (128 J/(kg·K)) ≈ 1.77

Both ratios are roughly equal. This implies that the number of atoms of each element in the mixture directly affects the specific heat of the alloy. In this case, it suggests that the specific heat of the alloy is influenced by both the specific heat and the relative proportion of each element.