If you don't practice enough it's obviously going to be hard but if you practice enough it's going to be a piece of cake so don't think if it's going to be hard or not just think it's going to be worth the try at the very end
I believe the answer you are looking for is the 4th one.
Answer:
81 °C
Explanation:
This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:
q = heat added/released by a sample
m = mass of sample
c=specific heat of sample
ΔT = change in temperature
from here we can rearrange the equation to state:
q/(mc) = ΔT
1200J/((20.0g)(4.2J/g•°C)) = ΔT
14°C = ΔT
If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.
