Calculate the energy required to heat 382.0 mg of water from 28.7 °C to 38.8 °C. Assume the specific heat capacity of water under these conditions is 4.18 J⋅g⁻¹⋅K⁻¹. Round your answer to 3 significant digits.

Question

High School

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Answer :

OliviaLunaGracy OliviaLunaGracy · Answer 1 · 2024-09-12T01:12:59-04:00

To calculate the energy required to heat the water, we can use the formula for heat transfer:

[tex]q = m \times c \times \Delta T[/tex]

Where:

[tex]q[/tex] is the heat energy (in joules, J) that is required.
[tex]m[/tex] is the mass of the water (in grams, g). Since the question provides the mass in milligrams (mg), we need to convert it to grams: [tex]382.0 \ mg = 0.382 \ g[/tex].
[tex]c[/tex] is the specific heat capacity of water, which is given as [tex]4.18 \ \text{J}\cdot\text{g}^{-1}\cdot\text{K}^{-1}[/tex].
[tex]\Delta T[/tex] is the change in temperature (in Celsius, °C or Kelvin, K). Celsius and Kelvin can be interchangeable for temperature changes, so:
[tex]\Delta T = 38.8 \, ^\circ\text{C} - 28.7 \, ^\circ\text{C} = 10.1 \, ^\circ\text{C}[/tex]

Now, plug in these values:

[tex]q = 0.382 \, \text{g} \times 4.18 \, \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1} \times 10.1 \, ^\circ\text{C}[/tex]

[tex]q = 16.134636 \, \text{J}[/tex]

Rounding the answer to 3 significant digits, we get:

[tex]q \approx 16.1 \, \text{J}[/tex]

Therefore, the energy required to heat 382.0 mg of water from 28.7°C to 38.8°C is approximately 16.1 Joules.