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QuestionHow many kg of oil are needed to heat 190 kg190 \mathrm{~kg} of water from 27∘C27^{\circ} \mathrm{C} to 100∘C100^{\circ} \mathrm{C}?

Studdy Solution

STEP 1

Assumptions1. The mass of water to be heated is190 kg. The initial temperature of the water is27°C3. The final temperature of the water is100°C4. The specific heat of water is4.184 J/g°C5. The conversion factor from pounds to grams is1 lb =454 g6. The energy content of oil is45 MJ/kg7. The energy required to heat the water is provided entirely by the oil

STEP 2

First, we need to calculate the amount of energy required to heat the water. This can be done using the formulaQ=m⋅c⋅ΔQ = m \cdot c \cdot \Deltawhere- Q is the heat energy required (in Joules) - m is the mass of the water (in grams) - c is the specific heat of water (in J/g°C) - Δ is the change in temperature (in °C)

STEP 3

Convert the mass of the water from kilograms to grams.
m=190 kg=190,000 gm =190 \, kg =190,000 \, g

STEP 4

Calculate the change in temperature.
Δ=final−initial=100°C−27°C=73°C\Delta =_{final} -_{initial} =100°C -27°C =73°C

STEP 5

Now, plug in the values for the mass of the water, the specific heat of water, and the change in temperature to calculate the heat energy required.
Q=190,000 g⋅4.184 J/g°C⋅73°CQ =190,000 \, g \cdot4.184 \, J/g°C \cdot73°C

STEP 6

Calculate the heat energy required.
Q=190,000 g⋅4.184 J/g°C⋅73°C=5.8×109 JQ =190,000 \, g \cdot4.184 \, J/g°C \cdot73°C =5.8 \times10^9 \, J

STEP 7

Convert the heat energy required from Joules to MegaJoules.
Q=5.×109 J=5.×103 MJQ =5. \times10^9 \, J =5. \times10^3 \, MJ

STEP 8

Now, we can calculate the mass of oil required to provide this amount of energy. This can be done using the formulamoil=Qoilm_{oil} = \frac{Q}{_{oil}}where- m_oil is the mass of oil required (in kg) - Q is the heat energy required (in MJ) -_oil is the energy content of oil (in MJ/kg)

STEP 9

Plug in the values for the heat energy required and the energy content of oil to calculate the mass of oil required.
moil=5.8×3 MJ45 MJ/kgm_{oil} = \frac{5.8 \times^3 \, MJ}{45 \, MJ/kg}

STEP 10

Calculate the mass of oil required.
moil=5.8×103 MJ45 MJ/kg=129 kgm_{oil} = \frac{5.8 \times10^3 \, MJ}{45 \, MJ/kg} =129 \, kgTo two significant figures, the mass of oil required is130 kg.

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