when 15.0g of steam drops in temperature from 275.0 celsius to 250.0 celsius, how much heat energy is released?

q = mass steam x specific heat steam x (Tfinal-Tinitial)

Question, what is the specific heat of steam???

To determine the amount of heat energy released, you can use the equation:

Q = m × c × ΔT

where:
Q = heat energy released (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C, final temperature - initial temperature)

First, convert the mass of steam from grams to kilograms:
15.0 g = 0.015 kg

Next, determine the specific heat capacity of steam:
The specific heat capacity of steam is approximately 2.03 J/g°C.

Now, calculate the change in temperature:
ΔT = 250.0°C - 275.0°C = -25.0°C

Now, substitute the values into the equation:
Q = (0.015 kg) × (2.03 J/g°C) × (-25.0°C)

Q = -0.76425 J

Therefore, approximately 0.76425 joules of heat energy is released when 15.0g of steam drops in temperature from 275.0°C to 250.0°C.

To calculate the amount of heat energy released when steam drops in temperature, you need to use the heat equation:

Q = mcΔT

where:
Q is the heat energy released or absorbed (in joules),
m is the mass of the substance (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, you have the mass of the substance (15.0 g) and the change in temperature (ΔT = 275.0 °C - 250.0 °C = 25.0 °C).

The specific heat capacity of water is approximately 4.184 J/g·°C. Since steam is water in the gaseous state, we can use this value.

Plugging in the values:

Q = (15.0 g) × (4.184 J/g·°C) × (25.0 °C)
Q = 15.0 g × 4.184 J/g·°C × 25.0 °C
Q = 1563 J

Therefore, when 15.0 g of steam drops in temperature from 275.0 °C to 250.0 °C, approximately 1563 J of heat energy is released.