Using calories, how much heat will 32 grams of water absorb when it is heated from 25 C to 80 C?
q = heat absorbed = mass x specific heat water x (Tf - Ti) where
mass = 32 grams.
You will need to look up the specific heat of water. I think it's about 4 J/g*C.
Tf is final temperature. Ti is initial T.
Post your work if you get stuck.
To calculate the heat absorbed by water when it is heated, you can use the following formula:
Heat = mass * specific heat capacity * temperature change
Here's how to calculate it step by step:
1. Determine the mass of water:
Given that the mass of water is 32 grams.
2. Find the specific heat capacity of water:
The specific heat capacity of water is 1 calorie/gram °C.
3. Calculate the temperature change:
The temperature change is the final temperature minus the initial temperature.
Given that the initial temperature is 25 °C and the final temperature is 80 °C, the temperature change is (80 °C - 25 °C) = 55 °C.
4. Apply the formula:
Heat = mass * specific heat capacity * temperature change
Heat = (32 grams) * (1 calorie/gram °C) * (55 °C)
Finally, calculate the heat absorbed:
Heat = 32 * 1 * 55
Heat = 1,760 calories
Therefore, 32 grams of water will absorb 1,760 calories of heat when heated from 25 °C to 80 °C.