Question

A chunk of iron at 90.6°C was added to 200.0 g of water at 15.5°C. The specific heat of iron is 0.449 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.2°C. Assuming no heat was lost to the surroundings, what was the mass of iron added? [unit is g and right sig. figs]

Answer 1

To solve this problem, we can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the iron.

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

q_water = m_water * c_water * ΔT_water

where:
m_water is the mass of water (200.0 g)
c_water is the specific heat of water (4.18 J/g°C)
ΔT_water is the change in temperature of water (18.2°C - 15.5°C)

Now, let's calculate the heat gained by the water:

q_water = 200.0 g * 4.18 J/g°C * (18.2°C - 15.5°C)

The heat lost by the iron can be calculated using the same formula:

q_iron = m_iron * c_iron * ΔT_iron

where:
c_iron is the specific heat of iron (0.449 J/g°C)
ΔT_iron is the change in temperature of the iron (90.6°C - 18.2°C)

Now, let's calculate the heat lost by the iron:

q_iron = m_iron * 0.449 J/g°C * (90.6°C - 18.2°C)

Since the heat gained by the water is equal to the heat lost by the iron, we can set up an equation:

q_water = q_iron

200.0 g * 4.18 J/g°C * (18.2°C - 15.5°C) = m_iron * 0.449 J/g°C * (90.6°C - 18.2°C)

Now, we can solve for the mass of iron (m_iron):

m_iron = (200.0 g * 4.18 J/g°C * (18.2°C - 15.5°C)) / (0.449 J/g°C * (90.6°C - 18.2°C))

Calculating this expression will give us the mass of iron added, with the appropriate significant figures.