Can someone please explain how to do this problem? I am really confused. Thank you.
3. How much heat does 32.0 grams of water absorb when it is heated from 25.0 degrees Celsius to 80.0 degrees Celsius?
q = mass*specific heat water*(Tfinal-Tinitial) = ??.
mass is 32.0 g
Tfinal = 80.0
Tinitial = 25.0
You need to look up the specific heat of water and plug in the numbers. Solve for q. Post your work if you get stuck.
q = 32.0g (55.0)
q = 1760
So does q equal 1760
You haven't multiplied by the specific heat of water, or did you? And if so what are the units of your answer.
To solve this problem, you need to use the equation q = mass * specific heat * (Tfinal - Tinitial).
Given:
- mass = 32.0 grams
- Tfinal = 80.0 degrees Celsius
- Tinitial = 25.0 degrees Celsius
Step 1: Look up the specific heat of water
The specific heat of water is typically given as 4.18 J/g·°C. This value represents the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.
Step 2: Plug in the numbers
Now, substitute the values you have into the equation:
q = 32.0 g * 4.18 J/g·°C * (80.0°C - 25.0°C)
Step 3: Calculate
First, subtract the initial temperature from the final temperature:
q = 32.0 g * 4.18 J/g·°C * (55.0°C)
Next, multiply the mass by the specific heat and the temperature difference:
q = 32.0 g * 4.18 J/g·°C * 55.0°C
Finally, calculate the value:
q = 7273.6 J
The heat absorbed by the water is 7273.6 joules (J). Make sure to include the correct units in your answer.