High School

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

- Initial temperature: 20°C
- Final temperature: 80°C
- Heating time: 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 :

A- The specific heat of the sugar syrup can be converted to 2,060 kJ/kg·°C, and

b- the rate of heating, which is the heat energy transferred per unit time, is approximately 156.96054 kJ/s in SI units.

A- To convert the specific heat from Btul⁻¹F⁻¹ to kJkg⁻¹°C⁻¹, we use the conversion factor: 1 Btul⁻¹F⁻¹ = 4.184 kJkg⁻¹°C⁻¹.

Specific heat of the sugar syrup = 0.9 Btul⁻¹F⁻¹.

Converting to kJkg⁻¹°C⁻¹:

Specific heat = 0.9 Btul⁻¹F⁻¹ * 4.184 kJkg⁻¹°C⁻¹/Btul⁻¹F⁻¹

Specific heat ≈ 3.7584 kJkg⁻¹°C⁻¹.

(b) The rate of heating, or the heat energy transferred per unit time, can be calculated using the formula:

Rate of heating = (mass of the syrup) × (specific heat) × (temperature change) / (time)

The mass of the syrup can be calculated using the volume and density:

Mass = (volume) × (density) = 50 ft³ × 66.9 lb/ft³ ≈ 3345 lb ≈ 1516.05 kg.

The temperature change is ΔT = 80°C - 20°C = 60°C.

Plugging these values into the formula, we get:

Rate of heating = (1516.05 kg) × (0.388 kJ/(kg⋅°C)) × (60°C) / (30 min × 60 s/min) ≈ 156.96054 kJ/s.

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