Assume a piece of hot metal is added to a calorimeter containing 50mL of water at 22C. The amount of heat released by the metal was determined to be 1000J, and the equilibrium temperature in the calorimeter was 26C. Remembering that the specific heat of water is 4.18 J/gC, how much heat in joules does the calorimeter gain?
The answer is 164J. How do you get this? I tried Mass x specific heat x delta T and got 836 J but this is wrong.
See your post above.
To calculate the amount of heat gained by the calorimeter, you can use the formula:
Q = m * c * ΔT
where:
Q is the amount of heat gained or lost
m is the mass of the substance (in this case, the water) in grams
c is the specific heat capacity of the substance (in this case, water) in J/g°C
ΔT is the change in temperature of the substance (in this case, the water) in °C
Given:
m (mass of water) = 50mL = 50g (since 1 mL of water is equivalent to 1g)
c (specific heat of water) = 4.18 J/g°C
ΔT (change in temperature of water) = 26°C - 22°C = 4°C
Now substitute the values into the formula:
Q = 50g * 4.18 J/g°C * 4°C
Q = 836 J
Your calculation is correct; however, it seems that the answer provided (164J) is incorrect. The correct answer should indeed be 836J.
To find the amount of heat gained by the calorimeter, you can use the equation:
q = m * c * ΔT
where:
q is the heat 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 substance is water and the change in temperature (ΔT) is the difference between the final temperature (26°C) and the initial temperature (22°C). Let's calculate it step by step.
1. Calculate the mass of water in the calorimeter using its density:
The density of water is approximately 1 g/mL.
So, the mass of water in the calorimeter is
mass = volume * density = 50 mL * 1 g/mL = 50 g.
2. Now, plug the values into the equation:
q = m * c * ΔT
q = 50 g * 4.18 J/g°C * (26°C - 22°C)
q = 50 g * 4.18 J/g°C * 4°C
q = 836 J.
It seems you have correctly calculated the heat gained by the water in the calorimeter, which is 836 J. However, you are now looking to find the heat gained by the calorimeter itself. To do that, you need to consider that the heat gained by the metal (1000 J) is transferred to both the water and the calorimeter.
Since the total heat gained is 1000 J and the heat gained by the water is 836 J, the remaining heat is gained by the calorimeter:
Heat gained by calorimeter = Total heat gained - Heat gained by water
Heat gained by calorimeter = 1000 J - 836 J
Heat gained by calorimeter = 164 J
So, the heat gained by the calorimeter is 164 J.