If you used 2300 kcal you expend in energy in one day to heat 6.0 * 10^4g of water, what would its new temperature be
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
58.3
To find the new temperature of the water after it has been heated using 2300 kcal of energy, you can use the formula:
Q = mcΔT
where,
Q is the amount of heat energy transferred,
m is the mass of the water,
c is the specific heat capacity of water,
ΔT is the change in temperature.
To start, you know the amount of energy used is 2300 kcal. However, to use this value in the equation, it needs to be converted to units of joules (J) by multiplying it by 4.184 (1 kcal = 4.184 kJ).
2300 kcal * 4.184 kJ/kcal = 9611.2 kJ
The mass of the water is given as 6.0 * 10^4g, and the specific heat capacity of water is approximately 4.18 J/g°C.
Now, set up the equation using the known values:
9611.2 kJ = (6.0 * 10^4g) * (4.18 J/g°C) * ΔT
Simplify by canceling units and rearrange the equation to solve for ΔT:
ΔT = 9611.2 kJ / ((6.0 * 10^4g) * (4.18 J/g°C))
Multiply the mass of the water by the specific heat capacity:
ΔT = 9611.2 kJ / (250800 g * J/g°C)
Divide the energy by the product of the mass and specific heat capacity:
ΔT ≈ 0.01528°C
Therefore, the new temperature of the water would increase by approximately 0.01528°C after using 2300 kcal of energy.