If 74.5 grams of a hot metal was added to 56.4 grams of water and the temperature of the water rose from 20.4 to 33.4 degrees, how many KJ of energy was lost by the hot metal?
i need helping setting up where the numbers go. I'm lost
I set that up in your first post. I can try to find it if you can't.
Here it is.
http://www.jiskha.com/display.cgi?id=1300573867
To determine the amount of energy lost by the hot metal, you can use the heat equation:
q = m * c * ΔT
Where:
q is the amount of heat energy gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, the water is not the substance that gained or lost heat but acted as a calorimeter to measure the heat lost by the metal. Therefore, we need to use the specific heat capacity of water in the calculation.
The specific heat capacity of water is approximately 4.186 J/g°C. However, the problem requires the answer in kilojoules (KJ), so we need to convert the value to KJ/g°C by dividing it by 1000.
Now, let's set up the equation using the given values:
m (mass of the metal) = 74.5 g
ΔT (change in temperature) = 33.4°C - 20.4°C = 13.0°C
m (mass of water) = 56.4 g
c (specific heat capacity of water) = 4.186 J/g°C (or 0.004186 KJ/g°C)
Now, we can substitute the values into the equation:
q = m * c * ΔT
q = 74.5 g * 0.004186 KJ/g°C * 13.0°C
Now, you can simply multiply the three values together to calculate the energy lost by the hot metal in KJ.