Water at 29.0 °C is sprayed onto 0.188 kg of molten gold at 1063 °C (its melting point). The water boils away, forming steam at 100.0 °C and leaving solid gold at 1063 °C. What is the minimum mass of water that must be used?
wileyplus
To find the minimum mass of water that must be used, we need to calculate the heat absorbed by the water and compare it to the heat released by the gold during the phase change.
First, let's find the heat absorbed by the water during heating (from 29.0 °C to 100.0 °C).
We can use the formula:
Q = mcΔT
Where:
- Q is the heat absorbed/released
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
For water, the specific heat capacity is approximately 4186 J/kg°C.
So, the heat absorbed by the water during heating is:
Q_water = m_water * c_water * ΔT_water
= m_water * 4186 * (100.0 - 29.0)
Next, we find the heat released by the gold during cooling (from 1063 °C to 100.0 °C). Since the gold is in the molten state, the heat released can be calculated using the formula for the latent heat of fusion:
Q_gold = m_gold * L
Where:
- L is the latent heat of fusion for gold
- m_gold is the mass of gold
For gold, the latent heat of fusion is 63,000 J/kg.
So, the heat released by the gold during cooling is:
Q_gold = m_gold * 63000
To determine the minimum mass of water used, we need to find the point at which the heat absorbed by the water equals the heat released by the gold:
Q_water = Q_gold
m_water * 4186 * (100.0 - 29.0) = m_gold * 63000
Now, we substitute the given values:
m_water * 4186 * (100.0 - 29.0) = 0.188 * 63000
Next, solve the equation for m_water:
m_water = (0.188 * 63000) / (4186 * (100.0 - 29.0))
Calculating this expression gives us the minimum mass of water that must be used.
m_water = 4.967 kg
Therefore, the minimum mass of water needed is approximately 4.967 kg.