A 43 kg sample of water absorbs 302 kJ of heat.
If the water was initially at 29.8 degree celsius, what is its final temperature?
sorry about before i forgot the units.
q = mass water x specific heat water x delta T. I would convert heat to Joules and mass sample to grams to start.
To find the final temperature of the water, we can use the heat capacity formula:
Q = m * C * ΔT
Where:
Q - Heat absorbed or released by the substance (in joules)
m - Mass of the substance (in kilograms)
C - Specific heat capacity of the substance (in joules per kilogram per degree Celsius)
ΔT - Change in temperature (in degree Celsius)
In this case, we are given:
Q = 302,000 joules (since 1 kJ = 1000 J)
m = 43 kilograms
C = specific heat capacity of water (approximately 4.186 J/g°C or 4.186 kJ/kg°C)
ΔT = final temperature - initial temperature
We need to convert the mass from kilograms to grams and the temperature to Celsius:
m = 43 kg * 1000 g/kg = 43,000 g
ΔT = final temperature - 29.8°C
Now, let's rearrange the formula to solve for the final temperature:
ΔT = Q / (m * C)
Substituting in the given values:
ΔT = 302,000 J / (43,000 g * 4.186 J/g°C)
ΔT ≈ 1.45°C
Now we can solve for the final temperature:
Final Temperature = 29.8°C + ΔT
Final Temperature ≈ 31.25°C
Therefore, the final temperature of the water is approximately 31.25 degrees Celsius.