If you used the 2600 you expend in energy in one day to heat 7.0×104 of water at 20, what would be its new temperature?
I'm going to do some assuming here.
2600= 2,600 cal
7.0×10^4=7.0×10^4 g
and 20=20ºC
Use the following formula:
q=mc∆T
Where
q=2,600 cal
m=7.0×10^4 g
c= 1.00 cal/g °C
∆T=Tf-Ti=Tf-20ºC
solve for Tf
q/mc=Tf-20ºC
{2,600 cal/[(7.0×10^4 g)*(1.00 cal/g°C)]}+20ºC=Tf
*****if 2,600 is kcal, just change 2.6 x 10^6 to get the correct answer.
Note should say if 2,600 is kcal, just change it to 2.6 x 10^6 to get the correct answer.
To find the new temperature of water after heating it with a given amount of energy, we can use the formula:
Q = m * C * ΔT
Where:
Q is the amount of energy transferred (in Joules)
m is the mass of the water (in kilograms)
C is the specific heat capacity of water (approximately 4200 J/kg°C)
ΔT is the change in temperature
In this case, we are given the initial temperature of the water (20°C), the energy expended (2600 J), and the mass of the water (7.0×10^4 kg). We need to find the change in temperature, ΔT.
Rearranging the formula, we have:
ΔT = Q / (m * C)
Now let's substitute the given values into the formula:
ΔT = 2600 J / (7.0×10^4 kg * 4200 J/kg°C)
Calculating that expression gives us the change in temperature (ΔT).