Question

The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.07 kg of this oil from 23 °C to 191 °C?
From q to J

Answer 1

Answer:

\large \boxed{\textbf{609 kJ}}  

Explanation:

The formula for the heat absorbed is

q = mCΔT

Data:

m = 2.07 kg

T₁ = 23 °C

T₂ = 191 °C

C = 1.75 J·°C⁻¹g⁻¹

Calculations:

1. Convert kilograms to grams

2.07 kg = 2070 g

2. Calculate ΔT

ΔT = T₂ - T₁ = 191 - 23  = 168 °C

3. Calculate q

$q = \text{2070 g} \times 1.75 \text{ J ^{\circ} C ^{-1} g ^{-1} } \times 168 \,^{\circ}\text{C} = \text{609 000 J} = \textbf{609 kJ}\\\text{The heat energy required is }\large \boxed{\textbf{609 kJ}}$