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?
q = mass water x specific heat water x (Tfinal-Tinitial)
i need help setting this problem up. i don't understand which numbers go where
There is no setting up. Just substitute the numbers into the equation.
q = your answer.
mass water = 56.4g in the problem.
specific heat water = look this up in your text or notes. It is 4.184 J/gram I think.
Tfinal = 33.4 from the problem.
Tinitial = 20.4 from the problem.
q will be in joules, the problem asks for kJ; therefore, divide the answer you get by 1000 to convert J to kJ.
To calculate the amount of energy lost by the hot metal, we need to use the specific heat capacity equation:
Q = m * c * ΔT
Where:
Q is the energy
m is the mass
c is the specific heat capacity of the substance
ΔT is the change in temperature
For the water, we will use the specific heat capacity of water, which is approximately 4.18 J/g°C.
First, let's calculate the energy gained by the water:
m_water = 56.4 grams
c_water = 4.18 J/g°C
ΔT_water = 33.4°C - 20.4°C = 13°C (the rise in temperature of the water)
Q_water = m_water * c_water * ΔT_water
Now, let's calculate the energy lost by the hot metal:
m_metal = 74.5 grams
c_metal = ? (We need to look up the specific heat capacity of the metal)
Since we don't have the specific heat capacity of the metal, we cannot calculate the exact energy lost at the moment. Different metals have different specific heat capacities.
To calculate the energy lost, we need to know the specific heat capacity of the metal. Once we know the specific heat capacity, we can use the same equation: Q = m * c * ΔT.
You can find the specific heat capacity of the metal by looking up a table or searching online for the specific heat capacity value of that particular metal.