QuestionA metal was heated and placed in 100.0 mL of water at 21.2°C, reaching 32.0°C. Find the energy absorbed by the water. $\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}} \\ \mathrm{q}_{\mathrm{H}_{2} \mathrm{O}}=[?] \mathrm{J}$

Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. The initial temperature of the water is21. °C3. The final equilibrium temperature of the water is32.0 °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 calculate the mass of the water. We can do this by multiplying the volume of the water by the density of water.
$Mass_{H2} = Volume_{H2} \\times Density_{H2}$

STEP 3

Now, plug in the given values for the volume of water and the density of water to calculate the mass.
$Mass_{H2} =100.0\, mL \\times1\, g/mL$

STEP 4

Calculate the mass of the water.
$Mass_{H2} =100.0\, mL \\times1\, g/mL =100.0\, g$

STEP 5

Now, we need to calculate the change in temperature of the water. We can do this 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.
$\Delta =32.0^{\circ}C -21.2^{\circ}C$

STEP 7

Calculate the change in temperature.
$\Delta =32.0^{\circ}C -21.2^{\circ}C =10.^{\circ}C$

STEP 8

Now, we can calculate the energy absorbed by the water. We can do this by multiplying the mass of the water, the specific heat capacity of water, and the change in temperature.
$q_{H2} = Mass_{H2} \\times C_{H2} \\times \Delta$

STEP 9

Now, plug in the values for the mass of the water, the specific heat capacity of water, and the change in temperature to calculate the energy absorbed.
$q_{H2} =100.\, g \\times4.18\, J/g°C \\times.8°C$

STEP 10

Calculate the energy absorbed by the water.
$q_{H2} =100.0\, g \\times4.18\, J/g°C \\times10.8°C =4514.4\, J$ The water absorbed4514.4 J of energy.

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QuestionA metal was heated and placed in 100.0 mL of water at 21.2°C, reaching 32.0°C. Find the energy absorbed by the water. $\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}} \\ \mathrm{q}_{\mathrm{H}_{2} \mathrm{O}}=[?] \mathrm{J}$

Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. The initial temperature of the water is21. °C3. The final equilibrium temperature of the water is32.0 °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 calculate the mass of the water. We can do this by multiplying the volume of the water by the density of water.
$Mass_{H2} = Volume_{H2} \\times Density_{H2}$

STEP 3

Now, plug in the given values for the volume of water and the density of water to calculate the mass.
$Mass_{H2} =100.0\, mL \\times1\, g/mL$

STEP 4

Calculate the mass of the water.
$Mass_{H2} =100.0\, mL \\times1\, g/mL =100.0\, g$

STEP 5

Now, we need to calculate the change in temperature of the water. We can do this 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.
$\Delta =32.0^{\circ}C -21.2^{\circ}C$

STEP 7

Calculate the change in temperature.
$\Delta =32.0^{\circ}C -21.2^{\circ}C =10.^{\circ}C$

STEP 8

Now, we can calculate the energy absorbed by the water. We can do this by multiplying the mass of the water, the specific heat capacity of water, and the change in temperature.
$q_{H2} = Mass_{H2} \\times C_{H2} \\times \Delta$

STEP 9

Now, plug in the values for the mass of the water, the specific heat capacity of water, and the change in temperature to calculate the energy absorbed.
$q_{H2} =100.\, g \\times4.18\, J/g°C \\times.8°C$

STEP 10

Calculate the energy absorbed by the water.
$q_{H2} =100.0\, g \\times4.18\, J/g°C \\times10.8°C =4514.4\, J$ The water absorbed4514.4 J of energy.

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Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. The initial temperature of the water is21. °C3. The final equilibrium temperature of the water is32.0 °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 calculate the mass of the water. We can do this by multiplying the volume of the water by the density of water.MassH2=VolumeH2timesDensityH2Mass_{H2} = Volume_{H2} \\times Density_{H2}MassH2​=VolumeH2​timesDensityH2​

STEP 3

Now, plug in the given values for the volume of water and the density of water to calculate the mass.MassH2=100.0 mLtimes1 g/mLMass_{H2} =100.0\, mL \\times1\, g/mLMassH2​=100.0mLtimes1g/mL

STEP 4

Calculate the mass of the water.MassH2=100.0 mLtimes1 g/mL=100.0 gMass_{H2} =100.0\, mL \\times1\, g/mL =100.0\, gMassH2​=100.0mLtimes1g/mL=100.0g

STEP 5

Now, we need to calculate the change in temperature of the water. We can do this 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=32.0∘C−21.2∘C\Delta =32.0^{\circ}C -21.2^{\circ}CΔ=32.0∘C−21.2∘C

STEP 7

Calculate the change in temperature.Δ=32.0∘C−21.2∘C=10.∘C\Delta =32.0^{\circ}C -21.2^{\circ}C =10.^{\circ}CΔ=32.0∘C−21.2∘C=10.∘C

STEP 8

Now, we can calculate the energy absorbed by the water. We can do this by multiplying the mass of the water, the specific heat capacity of water, and the change in temperature.qH2=MassH2timesCH2timesΔq_{H2} = Mass_{H2} \\times C_{H2} \\times \DeltaqH2​=MassH2​timesCH2​timesΔ

STEP 9

Now, plug in the values for the mass of the water, the specific heat capacity of water, and the change in temperature to calculate the energy absorbed.qH2=100. gtimes4.18 J/g°Ctimes.8°Cq_{H2} =100.\, g \\times4.18\, J/g°C \\times.8°CqH2​=100.gtimes4.18J/g°Ctimes.8°C

STEP 10

Calculate the energy absorbed by the water.qH2=100.0 gtimes4.18 J/g°Ctimes10.8°C=4514.4 Jq_{H2} =100.0\, g \\times4.18\, J/g°C \\times10.8°C =4514.4\, JqH2​=100.0gtimes4.18J/g°Ctimes10.8°C=4514.4JThe water absorbed4514.4 J of energy.

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First, we need to calculate the mass of the water. We can do this by multiplying the volume of the water by the density of water.
$Mass_{H2} = Volume_{H2} \\times Density_{H2}$

Now, plug in the given values for the volume of water and the density of water to calculate the mass.
$Mass_{H2} =100.0\, mL \\times1\, g/mL$

Calculate the mass of the water.
$Mass_{H2} =100.0\, mL \\times1\, g/mL =100.0\, g$

Now, we need to calculate the change in temperature of the water. We can do this 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.
$\Delta =32.0^{\circ}C -21.2^{\circ}C$

Calculate the change in temperature.
$\Delta =32.0^{\circ}C -21.2^{\circ}C =10.^{\circ}C$

Now, we can calculate the energy absorbed by the water. We can do this by multiplying the mass of the water, the specific heat capacity of water, and the change in temperature.
$q_{H2} = Mass_{H2} \\times C_{H2} \\times \Delta$

Now, plug in the values for the mass of the water, the specific heat capacity of water, and the change in temperature to calculate the energy absorbed.
$q_{H2} =100.\, g \\times4.18\, J/g°C \\times.8°C$

Calculate the energy absorbed by the water.
$q_{H2} =100.0\, g \\times4.18\, J/g°C \\times10.8°C =4514.4\, J$ The water absorbed4514.4 J of energy.