QuestionCalculate the energy absorbed by 100g of water with specific heat 4.18 J/g°C, initial temp 21.2°C, final temp 46.2°C.

Studdy Solution

STEP 1

Assumptions1. The mass of the water is100.0 g. The specific heat of the water is4.18 J/g°C3. The initial temperature (Ti) of the water is21.°C4. The final temperature (f) of the water is46.°C5. The energy absorbed by the water can be calculated using the formula $Q = m \cdot c \cdot \Delta$ where- Q is the energy absorbed, - m is the mass, - c is the specific heat, - and Δ is the change in temperature.

STEP 2

First, we need to calculate the change in temperature (Δ). This can be done by subtracting the initial temperature from the final temperature.
$\Delta =_f -_i$

STEP 3

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =46.2°C -21.2°C$

STEP 4

Calculate the change in temperature.
$\Delta =46.2°C -21.2°C =25.0°C$

STEP 5

Now that we have the change in temperature, we can calculate the energy absorbed by the water. Plug in the values for the mass, specific heat, and change in temperature into the formula.
$Q = m \cdot c \cdot \Delta$ $Q =100.0\,g \cdot4.18\,J/g°C \cdot25.0°C$

STEP 6

Calculate the energy absorbed by the water.
$Q =100.0\,g \cdot4.18\,J/g°C \cdot25.0°C =10,450\,J$ The water in this experiment absorbed10,450 Joules of energy.

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QuestionCalculate the energy absorbed by 100g of water with specific heat 4.18 J/g°C, initial temp 21.2°C, final temp 46.2°C.

Studdy Solution

STEP 1

STEP 2

First, we need to calculate the change in temperature (Δ). This can be done by subtracting the initial temperature from the final temperature.
$\Delta =_f -_i$

STEP 3

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =46.2°C -21.2°C$

STEP 4

Calculate the change in temperature.
$\Delta =46.2°C -21.2°C =25.0°C$

STEP 5

Now that we have the change in temperature, we can calculate the energy absorbed by the water. Plug in the values for the mass, specific heat, and change in temperature into the formula.
$Q = m \cdot c \cdot \Delta$ $Q =100.0\,g \cdot4.18\,J/g°C \cdot25.0°C$

STEP 6

Calculate the energy absorbed by the water.
$Q =100.0\,g \cdot4.18\,J/g°C \cdot25.0°C =10,450\,J$ The water in this experiment absorbed10,450 Joules of energy.

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QuestionCalculate the energy absorbed by 100g of water with specific heat 4.18 J/g°C, initial temp 21.2°C, final temp 46.2°C.

Studdy Solution

STEP 1

STEP 2

First, we need to calculate the change in temperature (Δ). This can be done by subtracting the initial temperature from the final temperature.Δ=f−i\Delta =_f -_iΔ=f​−i​

STEP 3

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.Δ=46.2°C−21.2°C\Delta =46.2°C -21.2°CΔ=46.2°C−21.2°C

STEP 4

Calculate the change in temperature.Δ=46.2°C−21.2°C=25.0°C\Delta =46.2°C -21.2°C =25.0°CΔ=46.2°C−21.2°C=25.0°C

STEP 5

STEP 6

Calculate the energy absorbed by the water.Q=100.0 g⋅4.18 J/g°C⋅25.0°C=10,450 JQ =100.0\,g \cdot4.18\,J/g°C \cdot25.0°C =10,450\,JQ=100.0g⋅4.18J/g°C⋅25.0°C=10,450JThe water in this experiment absorbed10,450 Joules of energy.

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First, we need to calculate the change in temperature (Δ). This can be done by subtracting the initial temperature from the final temperature.
$\Delta =_f -_i$

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature.
$\Delta =46.2°C -21.2°C$

Calculate the change in temperature.
$\Delta =46.2°C -21.2°C =25.0°C$

Calculate the energy absorbed by the water.
$Q =100.0\,g \cdot4.18\,J/g°C \cdot25.0°C =10,450\,J$ The water in this experiment absorbed10,450 Joules of energy.