The specific heat for liquid ethanol is 2.46 J/(g•°C). When 210 g of ethanol is cooled from 50 °C to 5 °C, the surrounding 7.80 × 103 g of air absorbs the heat. The specific heat of air is 1.01J/(g•°C)
You don't have a question. I assume you want to determine the change in T of the air.
heat lost by ethanol + heat gained by air = 0
[mass ethanol x specific heat ethanol x (Tfinal-Tinitial)] + [mass air x specific heat air x (Tfinal-Tinitial)] = 0
Substitute the values from the problem and solve for (Tfinal-Tinitial) air (or delta T = change in T).
0.656 C
The specific heat for liquid ethanol is 2.46 J/(g•°C). When 210 g of ethanol is cooled from 50 °C to 5 °C, the surrounding 7.80 × 103 g of air absorbs the heat. The specific heat of air is 1.01J/(g•°C). What is the change in air temperature, assuming all the heat released by the ethanol is absorbed by the air?
To determine the amount of heat absorbed by the air, you can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed (in Joules)
m is the mass of the substance (in grams)
c is the specific heat of the substance (in J/(g•°C))
ΔT is the change in temperature
For the ethanol:
m = 210 g
c = 2.46 J/(g•°C)
ΔT = 50 °C - 5 °C = 45 °C
Therefore, the amount of heat absorbed by the air is:
Q_ethanol = mcΔT = (210 g)(2.46 J/(g•°C))(45 °C) = 23,157 J
Now let's calculate the heat absorbed by the air using the same equation.
For the air:
m = 7.80 × 10^3 g
c = 1.01 J/(g•°C)
ΔT = 45 °C (since the air is in thermal equilibrium with the ethanol)
Q_air = mcΔT = (7.80 × 10^3 g)(1.01 J/(g•°C))(45 °C) = 352,890 J
Therefore, the amount of heat absorbed by the air is 352,890 J.
Note: This calculation assumes that there is no heat exchange with the surroundings other than between the ethanol and the air, and that the specific heat of ethanol and air remains constant over the temperature range.