Calculate the final temperature of 289 of water initially at 35 upon absorption of 20 of heat.
289 WHAT? 35 WHAT? 20 WHAT?
q = mass x specific heat x (Tfinal-Tinitial)
To calculate the final temperature of the water, we can use the equation:
Q = mcΔT
Where:
Q = Heat absorbed (in joules)
m = Mass of water (in grams)
c = Specific heat capacity of water (in joules per gram per degree Celsius)
ΔT = Change in temperature (in degrees Celsius)
In this case, we are given:
Q = 20 joules
m = 289 grams
c = specific heat capacity of water (which is approximately 4.184 joules per gram per degree Celsius)
Initial temperature (T1) = 35 degrees Celsius
We need to find the final temperature (T2).
Rearranging the equation, we have:
ΔT = Q / (mc)
Plugging in the given values, we can solve for ΔT:
ΔT = 20 joules / (289 grams * 4.184 joules/(gram * degree Celsius))
ΔT ≈ 0.016 degrees Celsius
To find the final temperature, we add the change in temperature to the initial temperature:
T2 = T1 + ΔT
T2 = 35 degrees Celsius + 0.016 degrees Celsius
T2 ≈ 35.016 degrees Celsius
Therefore, the final temperature of the water, after absorbing 20 joules of heat, is approximately 35.016 degrees Celsius.