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QuestionHow many kg of oil are needed to heat 190 kg of water from 28°C to 100°C? Use specific heat of water: 4.184 J/g°C.

Studdy Solution

STEP 1

Assumptions1. The initial temperature of the water is 28∘C28^{\circ} \mathrm{C} . The final temperature of the water is 100∘C100^{\circ} \mathrm{C}
3. The mass of the water is 190 kg190 \mathrm{~kg}
4. The specific heat of liquid water is 4.184 J/g∘C4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}
5. The conversion factor from pounds to grams is 1lb=454 g1 \mathrm{lb}=454 \mathrm{~g}
6. The energy required to heat the water is provided by the oil7. The energy content of the oil is not given, so we will assume it to be 1 kg1 \mathrm{~kg} of oil provides 45 MJ45 \mathrm{~MJ} of energy, a typical value for heating oil

STEP 2

First, we need to find the amount of energy required to heat the water. We can do this using the formula for heat energy, which is the product of the mass, the specific heat, and the change in temperature.
Q=m⋅c⋅ΔQ = m \cdot c \cdot \Delta

STEP 3

Now, plug in the given values for the mass, specific heat, and change in temperature to calculate the heat energy. Note that we need to convert the mass from kilograms to grams, as the specific heat is given in terms of grams.
Q=190,000 g⋅.184 J/g∘C⋅(100∘C−28∘C)Q =190,000 \mathrm{~g} \cdot.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C} \cdot (100^{\circ} \mathrm{C} -28^{\circ} \mathrm{C})

STEP 4

Calculate the heat energy.
Q=190,000 g⋅4.184 J/g∘C⋅72∘C=57,165,120,000 JQ =190,000 \mathrm{~g} \cdot4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C} \cdot72^{\circ} \mathrm{C} =57,165,120,000 \mathrm{~J}

STEP 5

Convert the heat energy from joules to megajoules, as the energy content of the oil is given in megajoules.
Q=57,165,120,000 J=57,165.12 MJQ =57,165,120,000 \mathrm{~J} =57,165.12 \mathrm{~MJ}

STEP 6

Now that we have the energy required to heat the water, we can find the amount of oil needed. This can be done by dividing the heat energy by the energy content of the oil.
Oil=Q/Energy content of oilOil = Q / Energy\, content\, of\, oil

STEP 7

Plug in the values for the heat energy and the energy content of the oil to calculate the amount of oil needed.
Oil=57,165.12 MJ/45 MJ/kgOil =57,165.12 \mathrm{~MJ} /45 \mathrm{~MJ/kg}

STEP 8

Calculate the amount of oil needed.
Oil=57,165.12 MJ/45 MJ/kg=1,270 kgOil =57,165.12 \mathrm{~MJ} /45 \mathrm{~MJ/kg} =1,270 \mathrm{~kg}Therefore, approximately1,270 kilograms of oil are needed to heat190 kilograms of water from28 degrees Celsius to100 degrees Celsius.

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