CHAPTER 5 DRILL
- C Use the equation Q = mc∆T.
Q = (0.04 kg)(0.50 kJ/kg • °C)(50°C − 20°C)=0.6 kJ = 600 J
- D Phase changes obey the equation Q = mL, where m is the mass of the sample
and L is the latent heat of transformation. In this case, we find that
- A First, let’s figure out how much heat is required to bring the water to its
boiling point.
Q = mc∆T = (0.1 kg)(4.186 kJ/kg • °C)(100°C − 20°C) = 33 kJ
Once the water reaches 100°C, any additional heat will be absorbed
and begin the transformation to steam. To completely vaporize the
sample requires
Q = mL = (0.1 kg)(2260 kJ/kg) = 226 kJ
Since the 20°C water absorbed only 100 kJ of heat, enough heat was
provided to bring the water to boiling, but not enough to completely
vaporize it (at which point, the absorption of more heat would begin
to increase the temperature of the steam). Thus, the water will reach
and remain at 100°C.
- D The equation for volume expansion is ∆V = βVi∆T, so
- B Because the gas is confined, n remains constant, and because we’re told the
volume is fixed, V remains constant as well. Since R is a universal constant,
