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In determining the average rate of heating of a tank of 20% sugar syrup, the following data is given:

- Initial temperature: 20°C
- Final temperature: 80°C
- Time taken: 30 minutes
- Volume of sugar syrup: 50 ft³
- Density of sugar syrup: 66.9 lb/ft³
- Specific heat of sugar syrup: 0.9 Btu/lb°F

Tasks:

(a) Convert the specific heat to kJ/kg°C.

(b) Determine the rate of heating, which is the heat energy transferred per unit time, in SI units (kJ/s).

Answer :

The specific heat of the sugar syrup in kJ kg-1 °C is 0.94955 kJ kg-1 °C. The rate of heating, in kJ S-1, is 2874.16892 kJ/min.

In order to convert the specific heat to kJ kg-1 °C, we need to use the following conversion:

1 Btu 1b-1°F = 1055.06 J kg-1 °C

Since 1 Btu = 1.05506 kJ, the specific heat in kJ kg-1 °C is 0.9 * 1.05506

= 0.94955 kJ kg-1 °C.

To determine the rate of heating in kJ S-1, we need to calculate the heat energy transferred in unit time. The heat energy transferred can be found using the formula:

Q = mcΔT

where Q is the heat energy transferred, m is the mass of the sugar syrup, c is the specific heat, and ΔT is the change in temperature.

First, we need to calculate the mass of the sugar syrup using the density and volume:

mass = density * volume

mass = 66.9 lb/ft3 * 50 ft3

= 3345 lb

Since 1 lb = 0.45359237 kg, the mass of the sugar syrup is 3345 * 0.45359237

= 1515.736 kg.

Substituting the values into the formula, we get:

Q = 1515.736 * 0.94955 * (80 - 20)

Q = 1515.736 * 0.94955 * 60

Q = 86225.0676 kJ

Finally, we can determine the rate of heating by dividing the heat energy transferred by the time:

Rate of heating = Q / time

Rate of heating = 86225.0676 kJ / 30 min

Rate of heating = 2874.16892 kJ/min

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