A 50.0 g silver spoon at 20.0 degrees Celsius is placed in a cup of coffee at 90.0 degrees C. How much heat does the spoon absorb from the coffee to reach a temp. of 89.0 degrees C?
Rememeber this equation. It, or a slight modification of it, will work all of these heat problems.
q = mc(Tf - Ti)
q is the heat lost or gained (depending upon the problem.
m is the mass, usually in grams.
c is the specific heat. If m is in grams then specific heat must be in cal/g*c or joules/g*c depending upon the unit you want for q.
Tf is the final T.
Ti is the initial T.
You will need to look up the specific heat of silver if it isn't given in the problem.
I hope this helps. Post your work if you get stuck.
193.2
Sure, I can help you with that! Let's use the equation you mentioned:
q = mc(Tf - Ti)
Given:
m = 50.0 g (mass of the silver spoon)
c = specific heat of silver (you'll need to look this up)
Tf = 89.0 degrees C (final temperature)
Ti = 20.0 degrees C (initial temperature)
We need to calculate q, which represents the heat absorbed by the spoon.
Now, I have a question for you before we proceed: Why did the spoon go to the coffee party? Because it heard there were lots of hot cups! Okay, let's continue.
First, we need to find the change in temperature (Tf - Ti). In this case, it would be:
Tf - Ti = 89.0°C - 20.0°C = 69.0°C
Now, the specific heat of silver is needed to calculate the heat absorbed. I'm not sure about this value, but let me check my clown database for you. Ah, here it is! The specific heat of silver is 0.235 cal/g°C.
Finally, we can plug all the values into the equation:
q = (50.0 g) * (0.235 cal/g°C) * (69.0°C)
And after doing the math, we get the answer. But before I reveal it to you, here's a quick knock-knock joke: Knock, knock.
Who's there?
Coffee.
Coffee who?
Coffee smells great in the morning, doesn't it?
Okay, the answer is... 809.25 calories! Keep in mind that this answer is in calories. If you prefer to use joules, just convert calories to joules by multiplying it by 4.184.
I hope this helps! Let me know if you have any more questions or if you'd like another joke!
To find the heat absorbed by the spoon, we can use the equation:
q = mc(Tf - Ti)
Given:
m (mass of spoon) = 50.0 g
c (specific heat of silver) = specific heat of silver (which needs to be looked up)
Tf (final temperature) = 89.0 degrees C
Ti (initial temperature) = 20.0 degrees C
Since the specific heat of silver is not given, we need to look it up. The specific heat of silver is approximately 0.235 J/g°C.
Now, let's calculate the heat absorbed by the spoon:
q = (50.0 g) x (0.235 J/g°C) x (89.0°C - 20.0°C)
q = 50.0 g x 0.235 J/g°C x 69.0°C
q ≈ 808.5 J
Therefore, the spoon absorbs approximately 808.5 J of heat from the coffee to reach a temperature of 89.0 degrees C.
To calculate the heat absorbed by the silver spoon, we can use the equation q = mc(Tf - Ti), where:
- q represents the heat absorbed or gained by the object
- m is the mass of the object, given as 50.0 grams
- c is the specific heat of the material, which needs to be looked up for silver
- Tf is the final temperature of the object, given as 89.0 degrees Celsius
- Ti is the initial temperature of the object, given as 20.0 degrees Celsius
First, let's find the specific heat of silver. You can look it up in a reliable source such as a textbook or an online database. The specific heat of silver is approximately 0.235 cal/g°C.
Now, we can substitute the given values into the equation:
q = (50.0 g) * (0.235 cal/g°C) * (89.0°C - 20.0°C)
Calculating the difference in temperatures:
q = (50.0 g) * (0.235 cal/g°C) * (69.0°C)
Now, we can simplify the equation:
q = 806.25 cal
Therefore, the silver spoon absorbs approximately 806.25 calories of heat from the coffee to reach a temperature of 89.0 degrees Celsius.
q = mcTc
m = 50.0 g = 0.050 kg
c = specific heat capacity of silver = 234.3 J kg-1 °C -1
Tc = Change in temperature = 89°C – 20°C = 69 °C
q = 0.050 kg * 234.3 J kg-1 °C -1* 69 °C
Heat gained (q) = 8.1*10^2 J.