11.0 g of aluminum at 200C and 20.0 g of copper are dropped into 54.0 of ethyl alcohol at 15C. The temperature quickly comes to 25.0. What was the initial temperature of the copper?

Look up the specific heats of Cu, Al and ethyl alcohol. Call them Cc, Ca and Ce.

The heat lost by the Al PLUS the heat lost by the Cu equals the heat gained by the alcohol.

Write that as an equation and solve for the initial T(alcohol). (Call it T)

11*Ca(200 - 25) + 20*Cc(T-25) = 54*Ce*(25 -15)

You do the rest.

To find the initial temperature of the copper, we can use the principle of conservation of energy. The heat lost by the aluminum and copper as they cool down will be equal to the heat gained by the ethyl alcohol.

The heat lost by the aluminum (Q_al) can be calculated using the formula:

Q_al = mcΔT_al

where:
m = mass of aluminum = 11.0 g
c = specific heat capacity of aluminum = 0.897 J/g°C (which is the amount of energy required to raise the temperature of 1 gram of aluminum by 1 degree Celsius)
ΔT_al = change in temperature of aluminum = (initial temperature of aluminum - final temperature of aluminum)

The heat lost by the copper (Q_cu) can be calculated in the same way:

Q_cu = mcΔT_cu

where:
m = mass of copper = 20.0 g
c = specific heat capacity of copper = 0.385 J/g°C (which is the amount of energy required to raise the temperature of 1 gram of copper by 1 degree Celsius)
ΔT_cu = change in temperature of copper = (initial temperature of copper - final temperature of copper)

The heat gained by the ethyl alcohol (Q_eth) can be calculated using the formula:

Q_eth = mcΔT_eth

where:
m = mass of ethyl alcohol = 54.0 g
c = specific heat capacity of ethyl alcohol = 2.46 J/g°C (which is the amount of energy required to raise the temperature of 1 gram of ethyl alcohol by 1 degree Celsius)
ΔT_eth = change in temperature of ethyl alcohol = (final temperature of ethyl alcohol - initial temperature of ethyl alcohol)

Since heat is conserved, we can set up the equation:

Q_al + Q_cu = Q_eth

Substituting the formulas for each type of material:

mcΔT_al + mcΔT_cu = mcΔT_eth

Simplifying the equation, we have:

ΔT_al + ΔT_cu = ΔT_eth

We already have the values for ΔT_eth (10.0°C), ΔT_al (200°C - 25°C = 175°C), and the mass of each material. We can now solve for ΔT_cu:

ΔT_cu = ΔT_eth - ΔT_al
ΔT_cu = 10.0°C - 175°C
ΔT_cu = -165.0°C

The negative value implies that the copper was initially at a temperature lower than the final temperature of the ethyl alcohol.

Therefore, the initial temperature of the copper was 25°C - 165°C = -140°C.