what will be the final temp of 500g of 21c waterbullets are dropped in? when 500g of 324c lead
To determine the final temperature of the water and lead mixture, we can use the principle of heat transfer, which states that the total heat gained by an object equals the total heat lost by another object.
In this case, we have two substances: water and lead. The heat lost by the lead will be equal to the heat gained by the water. We can calculate this using the formula:
\(Q = m \cdot c \cdot \Delta T\)
Where:
- Q is the heat transferred (in joules or calories)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in joules/gram °C or calories/gram °C)
- ΔT is the change in temperature (in °C)
First, let's calculate the heat lost by the lead:
\(Q_{\text{lead}} = m_{\text{lead}} \cdot c_{\text{lead}} \cdot \Delta T_{\text{lead}}\)
Given:
- m_lead = 500g (mass of lead)
- c_lead = specific heat capacity of lead = 0.13 J/g°C (approximately)
- ΔT_lead = final temperature - initial temperature = final temperature - 324°C
Next, let's calculate the heat gained by the water:
\(Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\)
Given:
- m_water = 500g (mass of water)
- c_water = specific heat capacity of water = 4.18 J/g°C (approximately)
- ΔT_water = final temperature - initial temperature = final temperature - 21°C
Since the heat lost by the lead is equal to the heat gained by the water, we can set the two equations equal to each other:
\(Q_{\text{lead}} = Q_{\text{water}}\)
\(m_{\text{lead}} \cdot c_{\text{lead}} \cdot \Delta T_{\text{lead}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\)
Now, we can solve for the final temperature (\(T_{\text{final}}\)) by rearranging the equation:
\(\Delta T_{\text{water}} = \frac{m_{\text{lead}} \cdot c_{\text{lead}} \cdot \Delta T_{\text{lead}}}{m_{\text{water}} \cdot c_{\text{water}}}\)
Finally, we add the initial temperature (21°C) to the change in temperature to calculate the final temperature:
\(T_{\text{final}} = 21°C + \Delta T_{\text{water}}\)
By substituting the given values into the equations, you can calculate the final temperature of the water and lead mixture.