what mass of water can be heated from 45 celcius to 88 celcius with 28kj?
28,000 J = mass x specific heat x delta T.
To solve this problem, we can use the formula:
q = m * c * ΔT
where:
q is the amount of heat absorbed or released (in joules),
m is the mass of the substance being heated (in grams),
c is the specific heat capacity of the substance (in J/g°C),
and ΔT is the change in temperature (in °C).
In this case, we want to find the mass of water (m) that can be heated from 45°C to 88°C using 28 kJ (28,000 J) of heat energy.
First, we need to determine the specific heat capacity (c) of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Now, let's rearrange the formula to solve for the mass (m):
m = q / (c * ΔT)
Substituting the given values:
m = 28,000 J / (4.18 J/g°C * (88°C - 45°C))
m = 28,000 J / (4.18 J/g°C * 43°C)
m = 28,000 J / (180.74 J/g)
m ≈ 154.95 g
Therefore, approximately 154.95 grams of water can be heated from 45°C to 88°C using 28 kJ of heat energy.