Question

PLEASE HELP!!!!

A foundry worker places a 5.78 kg sheet of cobalt at a temperature of 11°C on top of a 16.6 kg sheet of lead at 63°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the two sheets of metal.

Answer 1

• 35ºC

Step-by-step explanation:

You need the specif heat capacities of both cobalt and lead.

• Specific heat of cobalt: 0.42 J/g.ºC

• Specific heat of lead: 0.13 J/g.ºC

When the two sheets reach the thermal equilibrium their temperatures are equal.

You can use the equations for the thermal heat to find the equilibrium temperature:

Thermal heat released by the hot sheet, lead:

• Q = m × C × ΔT

• Q = 16.6 kg × 0.13J/g.ºC × (63ºC - T)

Thermal heat absorbed by the cold sheet, cobalt:

• Q = m × C × ΔT

• Q = 5.78 kg × 0.42J/g.ºC × (T - 11ºC)

Equal the two equations to solve for T:

• 16.6 kg × 0.13J/g.ºC × (63ºC - T) = 5.78kg × 0.42J/g.ºC × (T - 11ºC)

I remove the units for easier handling:

• 135.954 - 2.158T = 2.4276T - 26.7036

• 4.5856T = 162.6576

• T = 35.47ºC

Round to 2 significant figures: 35ºC ← answer