Question

a)Suba (m=72.6 kg) is skiing from a hill. She is standing at 28.5 m from the earth surface level.
i.Calculate Suba’s gravitational potential energy at the top.
ii.Calculate Suba’s kinetic energy at the top.
iii.Calculate total mechanical energy at the top.
iv.Calculate Suba’s kinetic energy and gravitational potential energy at the halfway down point from the top.
v.Calculate Suba’s kinetic energy at the bottom of the hill.
vi.Calculate Suba’s speed at the bottom of the hill.

b)The temperature of the water decreases to 0 °C in 10 minutes.
(specific heat capacity of water = 4200 J kg-1K–1)
Calculate,
i. the energy lost by the water when the temperature decreases to 0 °C.
ii. the average rate at which the water is losing energy, in J s–1.
iii. the time is taken for the water at 0 °C to turn completely into ice.
(The specific latent heat of fusion of water = 3.3 X 105 J Kg-1)

Answer 1

a) i) The gravitational energy can be calculated as $E_{gr}=mgh = 72.6\,\mathrm{kg}\cdot9.81\,\mathrm{N/kg}\cdot28.5\,\mathrm{m} = 20298\,\mathrm{J}.$

ii) When Suba is on the top of the hill, she is standing still, so the kinetic energy $E_{kin} = 0.$

iii) The total mechanical energy is equal to the sum of gravitational and kinetic energies, so $E_{tot}=E_{gr}+E_{kin} = 20298\,\mathrm{J} + 0 = 20298\,\mathrm{J}.$

iv) When Suba is at the halfway down point from the top, her height is $28.5:2 = 14.25\,\mathrm{m},$  so the potential energy will be $E_{gr} = mgh = 72.6\,\mathrm{kg}\cdot9.817,\mathrm{N/kg}\cdot14.25\,\mathrm{m} = 10149\,\mathrm{J}.$

The total mechanical energy is conserved, so the kinetic energy will be $E_{kin} = 20298\,\mathrm{J} - 10149\,\mathrm{J} = 10149\,\mathrm{J}.$

v) At the bottom of the hill the potential energy is equal to 0 because the height is equal to 0. So the kinetic energy will be equal to the total mechanical energy, $E_{kin} = 20298\,\mathrm{J}.$

vi) The kinetic energy is $E_{kin} = \dfrac{mv^2}{2},$ so $v = \sqrt{\dfrac{2E_{kin}}{m}} = 23.6\,\mathrm{m/s}.$

b) i) The energy that is lost can be calculated as $\Delta Q = mc_w\Delta T = m\cdot4200\cdot\Delta T = 4200\,\mathrm{J/kg/K}\cdot m\cdot\Delta T.$ Unfortunately, we don't know the mass and initial temperature, so we may write only the formula above.

ii) The average rate of energy lost is $P = \dfrac{\Delta Q}{\Delta t} = \dfrac{ 4200\,\mathrm{J/kg/K}\cdot m\cdot\Delta T}{600\,\mathrm{s}} = 7\,\mathrm{J/kg/s/K}\cdot m\cdot\Delta T.$

iii) The energy needed to turn the water into ice is

$\Delta Q_2 = \lambda_i \cdot m = 3.3\cdot10^5\,\mathrm{J/kg}\cdot m.$ To obtain the time we should write

$\Delta t_2 = \dfrac{\Delta Q_2}{\Delta Q}\cdot \Delta t = \dfrac{3.3\cdot10^5\,\mathrm{J/kg}\cdot m}{ 4200\,\mathrm{J/kg/K}\cdot m\cdot\Delta T} \cdot 600\,\mathrm{s} = \dfrac{3.3\cdot10^5}{4200\,\mathrm{1/K}\cdot\Delta T}\cdot600\,\mathrm{s}.$

If we know the initial temperature and mass of the water, we will substitute these data in the formulae above and get the numerical answers.