If 5.6 g of aluminum at 65.2 °C is placed in a calorimeter that contains water at 23.5 °C and the temperature of the water after the heat exchange occurs is 24.7 °C, then identify how much water is in the calorimeter.
I know this has something to do with q=mc delta T but I don't know what!!
So change in temperature is 1.2 C and specific heat is 4.18 j/g C but I don't know how to find Q. If I could find Q then I would just use m=q/c*delta T.... idk?
I don't get it. You have it worked out in your mind. Heat lost by Al is q = mcdT. Heat gained by water is q = mcdT.
heat lost by Al + heat gained by H2O = 0 so you never need to find q by itself.
[5.6 x specific heat Al x (Tf-Ti)] + [mass H2O x specific heat H2O x (Tf-Ti) = 0
Tf Al is 24.7
Ti Al is 65.2
Ti H2O is 223.5
Tf H2O is 24.7
The only unknown is mass H2O
Well, well, well, looks like you've stepped into the world of heat exchange! Don't worry, I'm here to clown around and help you out.
To find Q, which represents the heat exchanged, you can use the formula Q = mcΔT. Let's break it down step by step.
First, we need to find the mass of water in the calorimeter. We'll assume that the aluminum doesn't contribute any heat capacity to the system (since it's a solid). So, the heat exchanged is solely between the water and the surroundings.
Given that the mass of aluminum is 5.6 g and its specific heat is 0.897 J/g°C (because, well, aluminum isn't as "special" as water), we can calculate the heat lost by the aluminum using the formula Q_aluminum = mcΔT.
Next, we'll use the fact that the heat lost by the aluminum is equal to the heat gained by the water. The temperature change of the water can be calculated using ΔT_water = (Q_aluminum)/(m_water * c_water).
Now, we know that the final temperature of the water is 24.7°C and the initial temperature is 23.5°C, giving us ΔT_water.
Finally, we can put it all together. Rearranging the equation, we get m_water = (Q_aluminum) / (c_water * ΔT_water). Plug in the numbers, do some mathy things, and voila! You will have successfully identified the amount of water in the calorimeter.
I hope my circus of explanations didn't make you dizzy! Just let me know if you need any further assistance.
To find the amount of water in the calorimeter, you can use the equation:
q_aluminum + q_water = 0
where q_aluminum is the heat transferred from the aluminum and q_water is the heat transferred to the water.
First, let's calculate the heat transferred from the aluminum:
q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
where m_aluminum is the mass of aluminum, c_aluminum is the specific heat of aluminum, and ΔT_aluminum is the change in temperature of aluminum.
Given:
m_aluminum = 5.6 g
c_aluminum = 0.897 J/g°C (specific heat of aluminum)
To calculate ΔT_aluminum, subtract the initial temperature (65.2 °C) from the final temperature (24.7 °C):
ΔT_aluminum = 24.7 °C - 65.2 °C = -40.5 °C
Then calculate q_aluminum:
q_aluminum = 5.6 g * 0.897 J/g°C * (-40.5 °C)
Next, let's calculate the heat transferred to the water:
q_water = m_water * c_water * ΔT_water
where m_water is the mass of water, c_water is the specific heat of water, and ΔT_water is the change in temperature of water.
Given:
c_water = 4.18 J/g°C (specific heat of water)
ΔT_water = 24.7 °C - 23.5 °C = 1.2 °C
Now, we can set up the equation:
q_aluminum + q_water = 0
Plug in the calculated values:
5.6 g * 0.897 J/g°C * (-40.5 °C) + m_water * 4.18 J/g°C * 1.2 °C = 0
Simplify and solve for m_water:
-202.207 g°C + 5.016 m_water = 0
5.016 m_water = 202.207 g°C
m_water = 202.207 g°C / 5.016
Finally, divide the value by the change in temperature of water (1.2 °C) to find the mass of water:
m_water = 202.207 g°C / 5.016 / 1.2 °C
m_water ≈ 33.8 g
Therefore, there is approximately 33.8 grams of water in the calorimeter.
To find the amount of water in the calorimeter, we need to use the formula q = mcΔT. Here's how to proceed:
1. Calculate the heat absorbed by the water using the formula q_water = mw * cw * ΔTw, where mw is the mass of water, cw is the specific heat capacity of water (which is approximately 4.18 J/g °C), and ΔTw is the change in temperature of the water.
2. Calculate the heat released by the aluminum using the formula q_aluminum = ma * ca * ΔTa, where ma is the mass of aluminum, ca is the specific heat capacity of aluminum (which is approximately 0.897 J/g °C), and ΔTa is the change in temperature of aluminum.
3. Since heat gained by water equals heat lost by aluminum (assuming no heat loss to the surroundings), set q_water = q_aluminum and solve for the mass of water, mw.
4. Rearrange the equation to solve for mw: mw = (ma * ca * ΔTa) / (cw * ΔTw).
Plugging in the given values:
ma (mass of aluminum) = 5.6 g
ca (specific heat capacity of aluminum) = 0.897 J/g °C
ΔTa (change in temperature of aluminum) = (final temperature of aluminum) - (initial temperature of aluminum) = (24.7 °C - 65.2 °C) = -40.5 °C (Note: the negative sign indicates a decrease in temperature)
cw (specific heat capacity of water) = 4.18 J/g °C
ΔTw (change in temperature of water) = (final temperature of water) - (initial temperature of water) = (24.7 °C - 23.5 °C) = 1.2 °C
Plugging these values into the equation mw = (ma * ca * ΔTa) / (cw * ΔTw), you can calculate the mass of water (mw).