If that a chunk of potassium weighing 7.55 g is dropped into 400.0 g of water at
25.0 0C, what is the final temperature (in 0C) of the water if all the heat released is used to warm
the water
You know, I just don't have all of those thermodynamic tables memorized. So I need to know the kJ/mol for that reaction and make sure you tell me the equation too. It probably is 2K + 2H2O ==> H2 + 2KOH
its -196.49kJ/mol
To find the final temperature of the water when a chunk of potassium is dropped into it, we need to use the concept of heat transfer and the specific heat capacity of water.
The heat released by the potassium will be used to warm the water, so we can use the equation:
q = m * c * ΔT
where:
q is the heat transferred (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (4.18 J/g·°C)
ΔT is the change in temperature
First, let's calculate the heat released by the potassium. We know that the heat released will be equal to the heat absorbed by the water:
q_potassium = q_water
Using the equation for heat:
m_potassium * c_potassium * ΔT_potassium = m_water * c_water * ΔT_water
We know the mass of the potassium (7.55 g) and its specific heat capacity (0.757 J/g·°C). We also know the mass of water (400.0 g) and the specific heat capacity of water (4.18 J/g·°C). The initial temperature (25.0 °C) is assumed to be the same for both substances.
Plugging in the values, we have:
(7.55 g) * (0.757 J/g·°C) * ΔT_potassium = (400.0 g) * (4.18 J/g·°C) * ΔT_water
Simplifying the equation:
ΔT_water = (7.55 g * 0.757 J/g·°C * ΔT_potassium) / (400.0 g * 4.18 J/g·°C)
Now we can substitute the known values into the equation to find the change in temperature of the water.