If 1200 cal of heat are added to 50 g of water initially at a temperature of 15 oC, what is the final temperature of the water?
the energy needed to increase or decrease the temperature of a given substance is given by
Q = mc(T2-T1)
where
Q = heat (in J or cal)
m = mass (in g)
c = specific heat capacity (in J/g*K or cal/g*K)
T = temperature (in Kelvin)
**change in temperature can have unit degree Celsius or Kelvin,, thus you can use it interchangeably (what i mean is that the units for c can also be J/(g*C))
signs:
Q: (+) when heat is absorbed by substance
Q: (-) when heat is released from substance
for water, c = 1 cal/g*K.
substituting,
1200 = (50)(1)(T2 - 15)
1200 = 50*(T2) - 750
1950 = 50*(T2)
T2 = 39 deg C
hope this helps~ :)
To find the final temperature of the water, we can use the equation:
q = m * c * ΔT
Where:
q is the heat energy absorbed or released,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
First, we need to find the heat energy absorbed by the water. In this case, the heat energy added to the water is 1200 calories.
Next, we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 1 calorie/gram * oC.
Finally, we need to find the change in temperature. We can use the formula:
ΔT = q / (m * c)
Let's insert the known values into the equation:
ΔT = 1200 calories / (50 grams * 1 calorie/gram * oC) = 24 oC
So the final temperature of the water is the initial temperature plus the change in temperature:
Final temperature = 15 oC + 24 oC = 39 oC
Therefore, the final temperature of the water is 39 oC.