QuestionHow much energy did 100.0 mL of water absorb when heated from 23.7°C to 27.8°C? Use $\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. The initial temperature of water is23.7°C3. The final equilibrium temperature is27.8°C4. The specific heat capacity of water is4.18 J/g°C5. The density of water is approximately1 g/mL

STEP 2

First, we need to convert the volume of water to mass. Since the density of water is approximately1 g/mL, the mass of water is equal to the volume.
$Mass_{H2} = Volume_{H2}$

STEP 3

Now, plug in the given value for the volume of water to calculate the mass.
$Mass_{H2} =100.0\, mL$

STEP 4

Calculate the mass of water.
$Mass_{H2} =100.0\, mL =100.0\, g$

STEP 5

Next, we need to calculate the change in temperature. This can be done by subtracting the initial temperature from the final temperature.
$\Delta =_{final} -_{initial}$

STEP 6

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =27.8^{\circ}C -23.^{\circ}C$

STEP 7

Calculate the change in temperature.
$\Delta =27.^{\circ}C -23.7^{\circ}C =4.1^{\circ}C$

STEP 8

Now that we have the mass of water and the change in temperature, we can calculate the amount of energy absorbed by the water using the formula $Q = m \cdot C \cdot \Delta$ where- $Q$ is the heat energy absorbed, - $m$ is the mass of the substance, - $C$ is the specific heat capacity of the substance, and- $\Delta$ is the change in temperature.

STEP 9

Plug in the values for the mass of water, the specific heat capacity of water, and the change in temperature to calculate the heat energy absorbed.
$Q =100.\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C$

STEP 10

Calculate the heat energy absorbed by the water.
$Q =100.0\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C =1714.18\, J$ The water absorbed1714.18 Joules of energy.

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QuestionHow much energy did 100.0 mL of water absorb when heated from 23.7°C to 27.8°C? Use $\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. The initial temperature of water is23.7°C3. The final equilibrium temperature is27.8°C4. The specific heat capacity of water is4.18 J/g°C5. The density of water is approximately1 g/mL

STEP 2

First, we need to convert the volume of water to mass. Since the density of water is approximately1 g/mL, the mass of water is equal to the volume.
$Mass_{H2} = Volume_{H2}$

STEP 3

Now, plug in the given value for the volume of water to calculate the mass.
$Mass_{H2} =100.0\, mL$

STEP 4

Calculate the mass of water.
$Mass_{H2} =100.0\, mL =100.0\, g$

STEP 5

Next, we need to calculate the change in temperature. This can be done by subtracting the initial temperature from the final temperature.
$\Delta =_{final} -_{initial}$

STEP 6

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =27.8^{\circ}C -23.^{\circ}C$

STEP 7

Calculate the change in temperature.
$\Delta =27.^{\circ}C -23.7^{\circ}C =4.1^{\circ}C$

STEP 8

STEP 9

Plug in the values for the mass of water, the specific heat capacity of water, and the change in temperature to calculate the heat energy absorbed.
$Q =100.\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C$

STEP 10

Calculate the heat energy absorbed by the water.
$Q =100.0\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C =1714.18\, J$ The water absorbed1714.18 Joules of energy.

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QuestionHow much energy did 100.0 mL of water absorb when heated from 23.7°C to 27.8°C? Use CH2O=4.18Jg⋅∘C\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}CH2​O​=4.18g⋅∘CJ​.

Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. The initial temperature of water is23.7°C3. The final equilibrium temperature is27.8°C4. The specific heat capacity of water is4.18 J/g°C5. The density of water is approximately1 g/mL

STEP 2

First, we need to convert the volume of water to mass. Since the density of water is approximately1 g/mL, the mass of water is equal to the volume.MassH2=VolumeH2Mass_{H2} = Volume_{H2}MassH2​=VolumeH2​

STEP 3

Now, plug in the given value for the volume of water to calculate the mass.MassH2=100.0 mLMass_{H2} =100.0\, mLMassH2​=100.0mL

STEP 4

Calculate the mass of water.MassH2=100.0 mL=100.0 gMass_{H2} =100.0\, mL =100.0\, gMassH2​=100.0mL=100.0g

STEP 5

Next, we need to calculate the change in temperature. This can be done by subtracting the initial temperature from the final temperature.Δ=final−initial\Delta =_{final} -_{initial}Δ=final​−initial​

STEP 6

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.Δ=27.8∘C−23.∘C\Delta =27.8^{\circ}C -23.^{\circ}CΔ=27.8∘C−23.∘C

STEP 7

Calculate the change in temperature.Δ=27.∘C−23.7∘C=4.1∘C\Delta =27.^{\circ}C -23.7^{\circ}C =4.1^{\circ}CΔ=27.∘C−23.7∘C=4.1∘C

STEP 8

STEP 9

Plug in the values for the mass of water, the specific heat capacity of water, and the change in temperature to calculate the heat energy absorbed.Q=100. g⋅4.18 Jg⋅∘C⋅4. ∘CQ =100.\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}CQ=100.g⋅4.18g⋅∘CJ​⋅4.∘C

STEP 10

Calculate the heat energy absorbed by the water.Q=100.0 g⋅4.18 Jg⋅∘C⋅4. ∘C=1714.18 JQ =100.0\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C =1714.18\, JQ=100.0g⋅4.18g⋅∘CJ​⋅4.∘C=1714.18JThe water absorbed1714.18 Joules of energy.

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QuestionHow much energy did 100.0 mL of water absorb when heated from 23.7°C to 27.8°C? Use $\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

First, we need to convert the volume of water to mass. Since the density of water is approximately1 g/mL, the mass of water is equal to the volume.
$Mass_{H2} = Volume_{H2}$

Now, plug in the given value for the volume of water to calculate the mass.
$Mass_{H2} =100.0\, mL$

Calculate the mass of water.
$Mass_{H2} =100.0\, mL =100.0\, g$

Next, we need to calculate the change in temperature. This can be done by subtracting the initial temperature from the final temperature.
$\Delta =_{final} -_{initial}$

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =27.8^{\circ}C -23.^{\circ}C$

Calculate the change in temperature.
$\Delta =27.^{\circ}C -23.7^{\circ}C =4.1^{\circ}C$

Plug in the values for the mass of water, the specific heat capacity of water, and the change in temperature to calculate the heat energy absorbed.
$Q =100.\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C$

Calculate the heat energy absorbed by the water.
$Q =100.0\, g \cdot4.18\, \frac{J}{g \cdot ^{\circ}C} \cdot4.\, ^{\circ}C =1714.18\, J$ The water absorbed1714.18 Joules of energy.