Answer :
The final temperature [tex]\( T_{\text{final}} \)[/tex] of the system, where the lead and water reach thermal equilibrium, is approximately [tex]\( 49.04 \)[/tex] °C.
To find the final temperature when a piece of lead at 200 °C is inserted into water at 50 °C, we need to apply the principle of conservation of energy, assuming no heat loss to the surroundings.
Let's denote:
- [tex]\( m_L \)[/tex]: mass of lead (2 kg)
- [tex]\( T_L \)[/tex]: initial temperature of lead (200 °C)
- [tex]\( c'_L \)[/tex]: specific heat capacity of lead (128 J/kg°C)
- [tex]\( m_W \)[/tex]: mass of water (10 kg)
- [tex]\( T_W \)[/tex]: initial temperature of water (50 °C)
- [tex]\( c_W \)[/tex]: specific heat capacity of water (4184 J/kg°C)
The total heat lost by the lead will be equal to the heat gained by the water, assuming thermal equilibrium is reached.
The heat lost by the lead [tex]\( Q_{\text{lead}} \)[/tex] can be calculated using the formula:
[tex]\[ Q_{\text{lead}} = m_L \cdot c'L \cdot (T{\text{final}} - T_L) \][/tex]
The heat gained by the water [tex]\( Q_{\text{water}} \)[/tex] can be calculated using:
[tex]\[ Q_{\text{water}} = m_W \cdot c_W \cdot (T_{\text{final}} - T_W) \][/tex]
Since [tex]\( Q_{\text{lead}} = Q_{\text{water}} \)[/tex], we equate them:
[tex]\[ m_L \cdot c'L \cdot (T{\text{final}} - T_L) = m_W \cdot c_W \cdot (T_{\text{final}} - T_W) \][/tex]
Now, plug in the values:
[tex]\[ 2 \cdot 128 \cdot (T_{\text{final}} - 200) = 10 \cdot 4184 \cdot (T_{\text{final}} - 50) \][/tex]
Simplify and solve for [tex]\( T_{\text{final}} \)[/tex]:
[tex]\[ 256 \cdot (T_{\text{final}} - 200) = 41840 \cdot (T_{\text{final}} - 50) \][/tex]
Expand and solve the equation step by step:
[tex]\[ 256 \cdot T_{\text{final}} - 51200 = 41840 \cdot T_{\text{final}} - 2092000 \][/tex]
[tex]\[ 256 \cdot T_{\text{final}} - 41840 \cdot T_{\text{final}} = -2092000 + 51200 \][/tex]
[tex]\[ -41584 \cdot T_{\text{final}} = -2040800 \][/tex]
[tex]\[ T_{\text{final}} = \frac{2040800}{41584} \][/tex]
[tex]\[ T_{\text{final}} \approx 49.04 \][/tex]
Given that the mass of lead, m = 2kg and initial temperature, Tl= 200 deg C
and the mass of water, M = 10kg and initial temperature of water be Tw = 50 deg C
Also, the heat capacity of lead is, Cl = 128 J /kg deg C
and heat capacity of water is Cw = 4184 J /kg deg C
We have to find the final temperature of the lead and water.
As lead is immersed in water, temperature exchange occurs between water and lead until the temperature of lead and water becomes the same. this will be the final temperature, let this final temperature be denoted by T.
So,
[tex]\begin{gathered} \text{Heat of lead = Heat of water } \\ \end{gathered}[/tex]Let the heat of lead be Ql and the heat of water be Qw.
[tex]mC_l(T-T_l)=MC_w(T-T_w)[/tex]Substituting the values in the above equation,
[tex]\begin{gathered} 2\times128\times(T-200)=10\times4184\times(T-50) \\ 41840T-256T=\text{ 209200-51200} \\ T=\frac{2040800}{41584} \\ \\ =49.07^{\circ}C \end{gathered}[/tex]
