You have 50 grams of water that begins at a temperature of 28°C. If you add 100 calories of heat to this amount of water, what is the final temperature?
Group of answer choices
30°C
42°C
26°C
100°C
To calculate the final temperature, you can use the formula:
q = mcΔT
where:
q = heat added (100 calories)
m = mass of water (50 grams)
c = specific heat capacity of water (1 calorie/gram°C)
ΔT = change in temperature (Tf - Ti)
Rearranging the formula to solve for the final temperature (Tf):
ΔT = q / (m * c)
ΔT = 100 calories / (50g * 1 calorie/gram°C)
ΔT = 2°C
Now, since the initial temperature is 28°C and the temperature change is 2°C, we can calculate the final temperature:
Tf = Ti + ΔT
Tf = 28°C + 2°C
Tf = 30°C
Therefore, the final temperature of the water after adding 100 calories of heat is 30°C.