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Math  /  Numbers & Operations

QuestionCalculate the energy required to heat 0.30 kg of water from 21.9∘C21.9^{\circ} \mathrm{C} to 44.2∘C44.2^{\circ} \mathrm{C}. Assume the specific heat capacity of water under these conditions is 4.18 J⋅ g−1⋅ K−14.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}. Be sure your answer has the correct number of significant digits.

Studdy Solution

STEP 1

1. The mass of water is 0.30 kg 0.30 \, \text{kg} .
2. The initial temperature of water is 21.9∘C 21.9^{\circ} \text{C} .
3. The final temperature of water is 44.2∘C 44.2^{\circ} \text{C} .
4. The specific heat capacity of water is 4.18 J⋅g−1⋅K−1 4.18 \, \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1} .
5. The temperature change is the same in Celsius and Kelvin.

STEP 2

1. Convert mass from kilograms to grams.
2. Calculate the temperature change.
3. Use the formula for heat energy.
4. Calculate the energy required.
5. Ensure the answer has the correct number of significant digits.

STEP 3

Convert the mass of water from kilograms to grams:
0.30 kg=300 g 0.30 \, \text{kg} = 300 \, \text{g}

STEP 4

Calculate the temperature change (ΔT\Delta T):
ΔT=44.2∘C−21.9∘C=22.3 K \Delta T = 44.2^{\circ} \text{C} - 21.9^{\circ} \text{C} = 22.3 \, \text{K}

STEP 5

Use the formula for heat energy:
Q=m⋅c⋅ΔT Q = m \cdot c \cdot \Delta T
where: - Q Q is the heat energy, - m m is the mass in grams, - c c is the specific heat capacity, - ΔT \Delta T is the temperature change.

STEP 6

Substitute the known values into the formula:
Q=300 g⋅4.18 J⋅g−1⋅K−1⋅22.3 K Q = 300 \, \text{g} \cdot 4.18 \, \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1} \cdot 22.3 \, \text{K}
Calculate Q Q :
Q=300×4.18×22.3=27981.4 J Q = 300 \times 4.18 \times 22.3 = 27981.4 \, \text{J}

STEP 7

Ensure the answer has the correct number of significant digits. The least number of significant digits in the given data is three (from the mass, 0.30 kg0.30 \, \text{kg}).
Thus, the energy required is:
Q=2.80×104 J Q = 2.80 \times 10^4 \, \text{J}
The energy required to heat the water is:
2.80×104 J\boxed{2.80 \times 10^4 \, \text{J}}

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