QuestionHow much energy is needed to heat 620.0 g of copper from 14.25°C to 23.05°C if specific heat is 0.3850 J/(g°C)?

Studdy Solution

STEP 1

Assumptions1. The weight of the copper tubing is620.0 g. The initial temperature of the copper tubing is $14.25{ }^{\circ} \mathrm{C}$
3. The final temperature of the copper tubing is $23.05{ }^{\circ} \mathrm{C}$
4. The specific heat of copper is $0.3850 \mathrm{~J} /\left(\mathrm{g}\right.$ . ${ }^{\circ} \mathrm{C}$ )

STEP 2

We need to find the energy required to heat the copper tubing. We can use the formula for heat transfer, which is $q = m \cdot c \cdot \Delta$ where- $q$ is the heat energy, - $m$ is the mass of the substance, - $c$ is the specific heat of the substance, and- $\Delta$ is the change in temperature.

STEP 3

First, we need to calculate the change in temperature, which is the final temperature minus the initial temperature $\Delta =_{final} -_{initial}$

STEP 4

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature $\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C}$

STEP 5

Calculate the change in temperature $\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C} =8.8{ }^{\circ} \mathrm{C}$

STEP 6

Now that we have the change in temperature, we can substitute all the known values into the heat transfer formula to calculate the heat energy $q = m \cdot c \cdot \Delta$

STEP 7

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat energyq =620.0 \mathrm{~g} \cdot0.3850 \mathrm{~J} /\left(\mathrm{g}\right.$. ${ }^{\circ} \mathrm{C}$ ) \cdot.{ }^{\circ} \mathrm{C}

STEP 8

Calculate the heat energyq =620.0 \mathrm{~g} \cdot0.3850 \mathrm{~J} /\left(\mathrm{g}\right.$. ${ }^{\circ} \mathrm{C}$ ) \cdot8.8{ }^{\circ} \mathrm{C} =2090 \mathrm{~J}The energy required to heat the section of the copper tubing from $14.25{ }^{\circ} \mathrm{C}$ to $23.05{ }^{\circ} \mathrm{C}$ is $2090 \mathrm{~J}$ .

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QuestionHow much energy is needed to heat 620.0 g of copper from 14.25°C to 23.05°C if specific heat is 0.3850 J/(g°C)?

Studdy Solution

STEP 1

STEP 2

We need to find the energy required to heat the copper tubing. We can use the formula for heat transfer, which is $q = m \cdot c \cdot \Delta$ where- $q$ is the heat energy, - $m$ is the mass of the substance, - $c$ is the specific heat of the substance, and- $\Delta$ is the change in temperature.

STEP 3

First, we need to calculate the change in temperature, which is the final temperature minus the initial temperature $\Delta =_{final} -_{initial}$

STEP 4

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature $\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C}$

STEP 5

Calculate the change in temperature $\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C} =8.8{ }^{\circ} \mathrm{C}$

STEP 6

Now that we have the change in temperature, we can substitute all the known values into the heat transfer formula to calculate the heat energy $q = m \cdot c \cdot \Delta$

STEP 7

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat energyq =620.0 \mathrm{~g} \cdot0.3850 \mathrm{~J} /\left(\mathrm{g}\right.$. ${ }^{\circ} \mathrm{C}$ ) \cdot.{ }^{\circ} \mathrm{C}

STEP 8

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QuestionHow much energy is needed to heat 620.0 g of copper from 14.25°C to 23.05°C if specific heat is 0.3850 J/(g°C)?

Studdy Solution

STEP 1

STEP 2

STEP 3

First, we need to calculate the change in temperature, which is the final temperature minus the initial temperatureΔ=final−initial\Delta =_{final} -_{initial}Δ=final​−initial​

STEP 4

Now, plug in the given values for the initial and final temperatures to calculate the change in temperatureΔ=23.05∘C−14.25∘C\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C}Δ=23.05∘C−14.25∘C

STEP 5

Calculate the change in temperatureΔ=23.05∘C−14.25∘C=8.8∘C\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C} =8.8{ }^{\circ} \mathrm{C}Δ=23.05∘C−14.25∘C=8.8∘C

STEP 6

Now that we have the change in temperature, we can substitute all the known values into the heat transfer formula to calculate the heat energyq=m⋅c⋅Δq = m \cdot c \cdot \Deltaq=m⋅c⋅Δ

STEP 7

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat energyq =620.0 \mathrm{~g} \cdot0.3850 \mathrm{~J} /\left(\mathrm{g}\right.$. ${ }^{\circ} \mathrm{C}$ ) \cdot.{ }^{\circ} \mathrm{C}

STEP 8

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First, we need to calculate the change in temperature, which is the final temperature minus the initial temperature $\Delta =_{final} -_{initial}$

Now, plug in the given values for the initial and final temperatures to calculate the change in temperature $\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C}$

Calculate the change in temperature $\Delta =23.05{ }^{\circ} \mathrm{C} -14.25{ }^{\circ} \mathrm{C} =8.8{ }^{\circ} \mathrm{C}$

Now that we have the change in temperature, we can substitute all the known values into the heat transfer formula to calculate the heat energy $q = m \cdot c \cdot \Delta$