Question

What will be the equilibrium temperature when a 276 g block of copper at 280°C is placed in 821 g of water at 12.2°C?

Answer 1

We need to balance the energies.

Initially, the copper block has an internal energy of\[U_{c,i} = m_c c_c T_{c,i}\]and the water\[U_{w,i} = m_w c_w T_{w,i}\]where m is the mass, c is the specific heat and T_i is the initial temperature.

The sum of the initial energy must equal the sum of the final energy (assuming the heat exchange occurs in an adiabatic system, which is safe here).
\[m_c c_c T_{c,i} + m_w c_w T_{w,i} = m_c c_c T_{c,f} + m_w c_w T_{w,f}\]
Note that when the copper and water are in equilibrium, their temperatures will be equal. That is \(T_{c,f} = T_{w,f}\)\[T_f = {m_c c_c T_{c,i} + m_w c_w T_{w,i} \over m_c c_c + m_w c_w}\]

Alternatively, we could write that the change in energy of the copper must equal the change in energy of the water. This is the more typical expression, but sometimes harder to understand. This is expressed as\[m_c c_c (T_f - T_{c,i}) + m_w c_w (T_f - T_{w,i})=0\]Algebraic manipulation of this equation will yield the same expression for \(T_f\)

Answer 2

mc delta t = mcdelta t
0.821(4186)(T-12.2) = 0.276(387)(280-T)
3436.706(T-12.2) = 106.812(280-T)
3436.706T=41927.8132 = 29907.36-106.812T
3543.518T = 71835.1732
T=20.27227

Answer 3

It should be 0.276(387)(T-280)

Answer 4

Why?

Answer 5

I'm sorry. That is right. I thought you still had them summed together and equal to one.

Looks good.

Answer 6

Thank you!!