how many calories are lost when 15.0 ethanol,cools from 58.5 to -42.0
To calculate the calories lost when ethanol cools, we need to use the specific heat capacity and the change in temperature.
First, let's find the specific heat capacity of ethanol. The specific heat capacity (C) is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius (°C).
Ethanol has a specific heat capacity of 2.44 J/g°C. This means that it takes 2.44 Joules of energy to raise the temperature of 1 gram of ethanol by 1 degree Celsius.
Now, we need to calculate the mass of ethanol. You mentioned that there is 15.0 grams of ethanol.
Next, we need to calculate the change in temperature. The change in temperature is the final temperature (T₂) minus the initial temperature (T₁). In this case, the final temperature is -42.0°C, and the initial temperature is 58.5°C. So the change in temperature (ΔT) is (-42.0 - 58.5) = -100.5°C.
To calculate the calories lost, we first need to convert the change in temperature from Celsius to Kelvin. The conversion formula is K = °C + 273.15. So, we convert -100.5°C to Kelvin: -100.5 + 273.15 = 172.65 K.
Now we can calculate the calories lost using the equation: Calories lost = Specific heat capacity × Mass × Change in temperature.
Calories lost = 2.44 J/g°C × 15.0 g × 172.65 K
To convert the answer to calories, divide by 4.18 (since 1 calorie is equal to 4.18 joules).
Calories lost = (2.44 J/g°C × 15.0 g × 172.65 K) / 4.18
Finally, calculate the result: divide the numerator by the denominator.
Calories lost ≈ 1046.04 calories.