What mass of water in Gram can be heat from 300 K to 315 K by addition of 600 Joules of heat ?
600 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Thx!
To determine the mass of water that can be heated from 300 K to 315 K by adding 600 Joules of heat, we need to use the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 Joules per gram per Kelvin (J/g·K).
The formula for calculating the heat energy (Q) transferred to a substance is:
Q = m * c * ΔT
where:
Q is the heat energy (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g·K)
ΔT is the change in temperature (in Kelvin)
As mentioned, we want to find the mass of water (m), so we can rearrange the formula:
m = Q / (c * ΔT)
Plugging in the values:
Q = 600 J (given)
c = 4.18 J/g·K (specific heat capacity of water)
ΔT = 315 K - 300 K = 15 K
m = 600 J / (4.18 J/g·K * 15 K)
Finally, we can calculate the mass of water:
m ≈ 2.70 grams
Therefore, approximately 2.70 grams of water can be heated from 300 K to 315 K by adding 600 Joules of heat.