what is the change in temperature if 24g of water at 50 degrees C absorbs 985j of heat energy
q = mass H2O x specific heat H2O x delta T
To find the change in temperature, you can use the specific heat capacity equation:
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 (in J/g°C), and
ΔT is the change in temperature (in °C).
In this case:
m = 24g (mass of water)
c = 4.18 J/g°C (specific heat capacity of water)
Plugging in the values:
985 J = 24 g * 4.18 J/g°C * ΔT
Simplifying the equation:
ΔT = 985 J / (24 g * 4.18 J/g°C)
Calculating:
ΔT = 985 J / 100.32 g°C
ΔT ≈ 9.82 °C
Therefore, the change in temperature of 24g of water when it absorbs 985J of heat energy is approximately 9.82°C.
To determine the change in temperature of water when it absorbs a certain amount of heat energy, you can use the equation:
\(q = mcΔT\)
where:
- \(q\) is the heat energy absorbed or released by the substance (in joules, J)
- \(m\) is the mass of the substance (in grams, g)
- \(c\) is the specific heat capacity of the substance (in joules per gram per degree Celsius, J/g°C)
- \(ΔT\) is the change in temperature (in degrees Celsius, °C)
First, you need to find the specific heat capacity of water, which is approximately 4.18 J/g°C.
Next, you can substitute the given values into the equation:
\(985 J = (24 g) \cdot (4.18 J/g°C) \cdot ΔT\)
Now, you can solve the equation for ΔT:
\(ΔT = \frac{985 J}{(24 g) \cdot (4.18 J/g°C)}\)
Calculating this expression will give you the change in temperature of the water.