Water (2690 ) is heated until it just begins to boil. If the water absorbs 5.29×105 of heat in the process, what was the initial temperature of the water?express answer in Celsius.
q = m x Cg x (Tf - Ti)
where Tf=100 degree Celsius (boiling point of water)
find Ti.
http://www.ausetute.com.au/heatcapa.html
To find the initial temperature of the water, we need to use the concept of heat gained or lost by a substance. The formula is:
Q = mcΔT
Where:
Q = Heat gained or lost
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance
ΔT = Change in temperature
In this case, the water absorbs the heat, so it is gaining heat. We have the following information:
Q = 5.29×10^5 J (which is equal to calories since 1 calorie = 4.184 J)
m = Unknown
c = The specific heat capacity of water = 4.184 J/g°C
ΔT = The change in temperature = The final temperature - initial temperature
The final temperature of water at boiling point (when it just begins to boil) is 100°C.
Now we can rearrange the formula to solve for the initial temperature:
Q = mcΔT
Rearranging to isolate ΔT, we get:
ΔT = Q / (mc)
Substituting the known values:
ΔT = 5.29×10^5 J / (m * 4.184 J/g°C)
Since we want the temperature in Celsius, we need to use the same unit for mass. 1 mL of water has a mass of approximately 1g, so 2690 mL of water has a mass of 2690 grams.
ΔT = 5.29×10^5 J / (2690 g * 4.184 J/g°C)
Canceling out the units, we get:
ΔT = 5.29×10^5 / (2690 * 4.184)
Calculating this value:
ΔT ≈ 48.72 (rounded to two decimal places)
Finally, we can find the initial temperature by subtracting ΔT from the final temperature:
Initial temperature = final temperature - ΔT = 100°C - 48.72°C
Calculating this:
Initial temperature ≈ 51.28°C
Therefore, the initial temperature of the water was approximately 51.28°C.