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

Studdy Solution

STEP 1

Assumptions1. The initial volume of water is100.0 mL. . The initial temperature of the water is21.°C.
3. The final equilibrium temperature of the water is32.0°C.
4. The specific heat capacity of water (C) is4.18 J/g°C.
5. The density of water is approximately1 g/mL, so the mass of the water is approximately equal to its volume in mL.
6. The energy absorbed by the water (q) is what we need to find.

STEP 2

First, we need to find the change in temperature of the water. We can do this by subtracting the initial temperature from the final temperature.
$\Delta =_{final} -_{initial}$

STEP 3

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =32.0°C -21.2°C$

STEP 4

Calculate the change in temperature.
$\Delta =32.0°C -21.2°C =10.8°C$

STEP 5

Now that we have the change in temperature, we can find the energy absorbed by the water. The formula to calculate this is $q = m \cdot C \cdot \Delta$ where- m is the mass of the water- C is the specific heat capacity of the water- $\Delta$ is the change in temperature

STEP 6

Plug in the values for the mass of the water (which is approximately equal to its volume), the specific heat capacity, and the change in temperature to calculate the energy absorbed.
$q =100.0 \, g \cdot4.18 \, J/g°C \cdot10.8°C$

STEP 7

Calculate the energy absorbed by the water.
$q =100.0 \, g \cdot4.18 \, J/g°C \cdot10.°C =4514.4 \, J$ The water absorbed approximately4514.4 Joules 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. How much energy did the water absorb? $\mathrm{C}_{\mathrm{H}_{2} \mathrm{O}}=4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}} \\ \mathrm{q}_{\mathrm{H}_{2} \mathrm{O}}=[?] \mathrm{J}$

Studdy Solution

STEP 1

STEP 2

First, we need to find the change in temperature of the water. We can do this by subtracting the initial temperature from the final temperature.
$\Delta =_{final} -_{initial}$

STEP 3

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =32.0°C -21.2°C$

STEP 4

Calculate the change in temperature.
$\Delta =32.0°C -21.2°C =10.8°C$

STEP 5

Now that we have the change in temperature, we can find the energy absorbed by the water. The formula to calculate this is $q = m \cdot C \cdot \Delta$ where- m is the mass of the water- C is the specific heat capacity of the water- $\Delta$ is the change in temperature

STEP 6

Plug in the values for the mass of the water (which is approximately equal to its volume), the specific heat capacity, and the change in temperature to calculate the energy absorbed.
$q =100.0 \, g \cdot4.18 \, J/g°C \cdot10.8°C$

STEP 7

Calculate the energy absorbed by the water.
$q =100.0 \, g \cdot4.18 \, J/g°C \cdot10.°C =4514.4 \, J$ The water absorbed approximately4514.4 Joules of energy.

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Studdy Solution

STEP 1

STEP 2

First, we need to find 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 3

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°C -21.2°CΔ=32.0°C−21.2°C

STEP 4

Calculate the change in temperature.Δ=32.0°C−21.2°C=10.8°C\Delta =32.0°C -21.2°C =10.8°CΔ=32.0°C−21.2°C=10.8°C

STEP 5

Now that we have the change in temperature, we can find the energy absorbed by the water. The formula to calculate this isq=m⋅C⋅Δq = m \cdot C \cdot \Deltaq=m⋅C⋅Δwhere- m is the mass of the water- C is the specific heat capacity of the water- Δ\DeltaΔ is the change in temperature

STEP 6

Plug in the values for the mass of the water (which is approximately equal to its volume), the specific heat capacity, and the change in temperature to calculate the energy absorbed.q=100.0 g⋅4.18 J/g°C⋅10.8°Cq =100.0 \, g \cdot4.18 \, J/g°C \cdot10.8°Cq=100.0g⋅4.18J/g°C⋅10.8°C

STEP 7

Calculate the energy absorbed by the water.q=100.0 g⋅4.18 J/g°C⋅10.°C=4514.4 Jq =100.0 \, g \cdot4.18 \, J/g°C \cdot10.°C =4514.4 \, Jq=100.0g⋅4.18J/g°C⋅10.°C=4514.4JThe water absorbed approximately4514.4 Joules of energy.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

First, we need to find 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.
$\Delta =32.0°C -21.2°C$

Calculate the change in temperature.
$\Delta =32.0°C -21.2°C =10.8°C$

Now that we have the change in temperature, we can find the energy absorbed by the water. The formula to calculate this is $q = m \cdot C \cdot \Delta$ where- m is the mass of the water- C is the specific heat capacity of the water- $\Delta$ is the change in temperature

Plug in the values for the mass of the water (which is approximately equal to its volume), the specific heat capacity, and the change in temperature to calculate the energy absorbed.
$q =100.0 \, g \cdot4.18 \, J/g°C \cdot10.8°C$

Calculate the energy absorbed by the water.
$q =100.0 \, g \cdot4.18 \, J/g°C \cdot10.°C =4514.4 \, J$ The water absorbed approximately4514.4 Joules of energy.