A 45kg sample of water absorbs 367kJ of heat. If the water was initially 24.8ºC, what is final temperature? Please explain steps. Thanks
I don't appreciate the LORD: if Jesus had anything to do with it I would NOT have to take chemistry!
where did you find the C?
mass of water= 367,000 x specific heat (24.8?) x Tf (how do i find this? what part of the question tells me?)
To determine the final temperature of the water, you can use the equation:
Q = mcΔT
where:
Q is the heat absorbed by the substance (in joules)
m is the mass of the substance (in kilograms)
c is the specific heat capacity of the substance (in J/kg·ºC)
ΔT is the change in temperature (final temperature - initial temperature)
In this case, we are given the mass of the water (m = 45 kg) and the heat absorbed by the water (Q = 367 kJ = 367,000 J). The specific heat capacity of water (c) is 4.18 J/g·ºC.
To use the equation above, we need to convert the mass from grams to kilograms, and the heat from kilojoules to joules:
m = 45 kg
Q = 367,000 J
Now, we can rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
First, we need to convert the specific heat capacity to J/kg·ºC:
c = 4.18 J/g·ºC = 4180 J/kg·ºC (multiplied by 1000 to convert grams to kilograms)
Now substitute the values into the equation:
ΔT = 367,000 J / (45 kg * 4180 J/kg·ºC)
Calculating the value gives:
ΔT ≈ 1.98ºC
Lastly, to find the final temperature, we add ΔT to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 24.8ºC + 1.98ºC
Therefore, the final temperature of the water is approximately 26.78ºC.
heat=mass*c*(tf-ti)
steps? You know heat, mass, ti. The only issue is the specific heat of water UNITS. YOu need c in kJ/kgC
Lord:
mass water=45kg
heat=367kJ
c= 4.17kJ/kgC
Ti=24.8C
367kJ=45kg*4.17kJ/kgC*(Tf-20)
Tf-20=367/(45*4.17)
Tf=367/(45*4.17) +20