Aidan S. answered • 12/05/23
UCSB Grad for Science and Math Tutoring
This question is what I like to call a "combo problem" meaning that you have to use q=mCΔT and q=mΔH. You can identify this by looking at the change in temperature. Water starts at -10°C and goes through a phase change at 0 degrees and 100 degrees. At phase changes we must use q=mΔH and with temperature changes we use q=mCΔT.
To start, calculate the energy required to change the temperature from -10 to 0°C using q=mC(s)ΔT.
q is energy, m is mass, C is the specific heat of solid water, and ΔT is the change in temperature: Tfinal (0) - Tinitial (-10).
q=(75g)(2.087 J/g·°C)(0 - (-10) °C) = 1,565.25 J
Next, a phase change happens at 0°C which is solid -> liquid. This is called melting or fusion. Calculate the energy to change water from solid to liquid using q=mΔH. The correct ΔH value is given in your chart under "enthalpy of fusion."
q=(75g)(333.6 J/g) = 25,020 J
Next, calculate the energy required to change the water from 0 to 100°C using q=mC(l)ΔT. This time C is the specific heat of liquid water.
q=(75g)(4.184 J/g·°C)(100 - 0°C ) = 313,800 J
Next, water goes through a phase change at 100°C from liquid -> gas. This is called vaporization. Calculate the energy to change water from liquid to gas using q=mΔH. The correct ΔH value is in your table under "enthalpy of vaporization."
q=(75g)(2257 J/g) = 169,275 J
Almost there! Calculate the energy required to change the temperature from 100 to 133°C using q=mC(g)ΔT. C is the specific heat of gaseous water.
q=(75g)(2.000 J/g·°C)(133 - 100°C ) = 4,950 J
Now add all of your calculated q values together!
1,565.25 + 25,020 + 313,800 + 169,275 + 4,950 = 514,610.25 J
If your teacher wants you to convert to kJ (1000 J = 1 kJ) divide your answer by 1,000. Your final answer would be 514.61 kJ
