Answer:
Q total = 2204.49 kJ
Explanation:
The mass of ice over the windshield is
mass = density * volume
and
volume = area * width= 1 m² * 0.007 m = 0.007 m³
thus
mass = 917 kg/m³ * 0.007 m³ = 6.419 kg
the heat that must reach the ice in order to melt it is
1) sensible heat until melting temperature ( T m = 0°C , assuming that pressure =1 atm since the windshield is exposed to the atmosphere)
2) Latent heat to melt the ice at its melting temperature
1) Q sen = m ice * c ice * ( T final - T initial) = 6.419 kg * 2.108 kJ / Kg °C ( 0°C - (-4°C)) = 54.125 kJ
2) Q lat = m ice * L ice = 6.419 kg * 335 kJ/kg ) = 2150.265 kJ
therefore Q total = Q sen + Q lat = 54.125 kJ + 2150.265 kJ = 2204.49 kJ
this is the minimum amount of heat required , since we did not take into account the heat losses to the surroundings
Note :
- the specific heat of water and ice and also the latent heat were taken from tables.
