An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice cream, of mass 0.6kg at a temperature of 20 °C, is added to the container.
container
refrigeration
unit
(a) Calculate how much energy must be removed from the mixture to cool it to its
freezing point of -16°C.
(Specific heat capacity of ice cream = 2100 J kg 1 °C')

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is