Terms
- Heat capacity is the amount of heat required to raise the temperature of a body by 1 K (or 1°C)
- Specific heat capacity is the amount of heat required to raise the temperature of 1kg of the substance by 1 K (or 1°C)
- Latent heat of fusion of a solid substance is the heat energy needed to change it from solid to liquid state without any change in temperature
- Specific latent heat of fusion of a solid substance is the heat energy needed to change 1kg of it from solid to liquid state without any change in temperature
- SI unit: J/kg
- Latent heat of vaporisation of a substance is the heat energy needed to change it from liquid vapour state without any change in temperature
- Specific latent heat of vaporisation of a substance is the heat energy needed to change 1kg of it from liquid to vapour state without any change in temperature
- SI unit: J/kg
Heat Change Formula
- Q = mcθ
- m = mass (kg)
c = specific heat capacity (J kg-1 oC-1)
θ = temperature change (o)
- m = mass (kg)
Electric heater
Energy Supply, E = Pt
Energy Received, Q = mcθ
Energy Supplied, E = Energy Receive, Q
Pt = mcθ
E = electrical Energy (J or Nm)
P = Power of the electric heater (W)
t = time (in second) (s)
Q = Heat Change (J or Nm)
m = mass (kg)
c = specific heat capacity (J kg-1 oC-1)
θ = temperature change (o)
Mixing 2 liquids
Heat Gain by Liquid 1 = Heat Loss by Liquid 2
m1c1θ1 = m2c2θ2
m1 = mass of liquid 1
c1 = specific heat capacity of liquid 1
θ1 = temperature change of liquid 1
m2 = mass of liquid 2
c2 = specific heat capacity of liquid 2
θ2 = temperature change of liquid 2
Specific Latent Heat
Q = mL
Q = Heat Change (J or Nm)
m = mass (kg)
L = specific latent heat (J kg-1)
1. A mass of 0.20kg of water at 0°C is placed in a vessel of negligible heat capacity. An electric heater with an output of 24 W is placed in the water and switched on. When the temperature of the water reaches 12°C, the heater is switched off.
a) Calculate the time for which the heater is switched on. Assume that the heat capacity of water is 4200J/kgK
b) An ice cube of mass 0.020kg is added to the 0.20kg of water at 0°C in the same vessel and the heater is switched on. Assuming that all the ice is at 0°C, calculate how long it will take for the water to reach 12°C. Assume that the specific latent heat of fusion on ice is 0.34 MJ/kg
Solutions
E = mcθ
E = Pt
24 × t = 0.20 × 4200 × 12
t = 420
–> The heater is switched on for 420 s.
b) Heat absorbed by ice = Heat used to melt ice + Heat used to raise temperature of ice water from 0°C to 12°C
= (0.020 × 340000) + (0.040 × 4200 × 12)
= 8816 J
Time t = 8816/24
= 367 s
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