A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200 joules, what is the approximate final temperature of the water? 75 °C
78 °C
81 °C
87 °C

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.