What is he final temperature after 80.0 calories is absorbed by 10.0 grams of water at 25.0 *C?
To find the final temperature after absorbing 80.0 calories of heat energy, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q is the heat energy absorbed or released (in calories)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in calories per gram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
In this case, we need to find the final temperature (ΔT) after absorbing 80.0 calories of heat energy. The mass (m) of water is 10.0 grams and the initial temperature is 25.0 °C.
Now, let's rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
We can substitute the known values into the equation:
ΔT = 80.0 calories / (10.0 grams * c)
The specific heat capacity (c) of water is approximately 1 calorie per gram per degree Celsius.
ΔT = 80.0 calories / (10.0 grams * 1 calorie/gram/°C)
Now, we can simplify the units:
ΔT = 8.0 °C
Therefore, the final temperature is 25.0 °C + 8.0 °C = 33.0 °C.