A 375 g piece of iron was heated to 975 °C and then quenched in a bucket containing one gallon of water at 20 °C. What was the temperature of the water after the iron and the water came to thermal equilibrium?

See your other post above.

To solve this problem, we need to apply the principles of heat transfer. The heat gained by the water is equal to the heat lost by the iron. We can use the formula:

Q = mcΔT

Where:
Q is the amount of heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

For the iron, the heat lost is given by:
Q_iron = m_iron * c_iron * ΔT_iron

For the water, the heat gained is given by:
Q_water = m_water * c_water * ΔT_water

Since the iron and water come to thermal equilibrium, the heat lost by the iron is equal to the heat gained by the water, so:

Q_iron = Q_water

Now let's plug in the given values and solve for ΔT_water.

Given:
m_iron = 375 g
c_iron (specific heat capacity of iron) = 0.45 J/g°C
ΔT_iron = 975°C - 20°C = 955°C

m_water (1 gallon of water) = 3785.41 g
c_water (specific heat capacity of water) = 4.18 J/g°C

Q_iron = Q_water

m_iron * c_iron * ΔT_iron = m_water * c_water * ΔT_water

375 g * 0.45 J/g°C * 955°C = 3785.41 g * 4.18 J/g°C * ΔT_water

Simplifying the equation:

ΔT_water = (375 g * 0.45 J/g°C * 955°C) / (3785.41 g * 4.18 J/g°C)

Calculating the result:

ΔT_water ≈ 22.8 °C

Therefore, the temperature of the water after the iron and water come to thermal equilibrium is approximately 22.8 °C.