QuestionCalculate the joules needed to heat 32.1 g of aluminum from $21.0^{\circ} \mathrm{C}$ to $29.5^{\circ} \mathrm{C}$ using $0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ .

Studdy Solution

STEP 1

Assumptions1. The specific heat of aluminum is $0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ . The mass of aluminum is32.1 grams3. The initial temperature is $21.0^{\circ} \mathrm{C}$
4. The final temperature is $29.5^{\circ} \mathrm{C}$
5. The formula to calculate heat is $q = mc\Delta$ , where $q$ is the heat, $m$ is the mass, $c$ is the specific heat, and $\Delta$ is the change in temperature

STEP 2

First, we need to calculate the change in temperature. 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.
$\Delta =29.5^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C}$

STEP 4

Calculate the change in temperature.
$\Delta =29.^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C} =8.^{\circ} \mathrm{C}$

STEP 5

Now that we have the change in temperature, we can calculate the amount of heat required to heat the aluminum. We can do this by using the formula $q = mc\Delta$ .
$q = m \times c \times \Delta$

STEP 6

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat.
$q =32.1 \mathrm{~g} \times0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times8.5^{\circ} \mathrm{C}$

STEP 7

Calculate the heat.
$q =32.1 \mathrm{~g} \times0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times.5^{\circ} \mathrm{C} =244.3 \mathrm{~J}$ The amount of heat required to heat32.1 grams of aluminum from $21.0^{\circ} \mathrm{C}$ to $29.5^{\circ} \mathrm{C}$ is244.3 Joules.

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QuestionCalculate the joules needed to heat 32.1 g of aluminum from $21.0^{\circ} \mathrm{C}$ to $29.5^{\circ} \mathrm{C}$ using $0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ .

Studdy Solution

STEP 1

STEP 2

First, we need to calculate the change in temperature. 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.
$\Delta =29.5^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C}$

STEP 4

Calculate the change in temperature.
$\Delta =29.^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C} =8.^{\circ} \mathrm{C}$

STEP 5

Now that we have the change in temperature, we can calculate the amount of heat required to heat the aluminum. We can do this by using the formula $q = mc\Delta$ .
$q = m \times c \times \Delta$

STEP 6

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat.
$q =32.1 \mathrm{~g} \times0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times8.5^{\circ} \mathrm{C}$

STEP 7

Calculate the heat.
$q =32.1 \mathrm{~g} \times0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times.5^{\circ} \mathrm{C} =244.3 \mathrm{~J}$ The amount of heat required to heat32.1 grams of aluminum from $21.0^{\circ} \mathrm{C}$ to $29.5^{\circ} \mathrm{C}$ is244.3 Joules.

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QuestionCalculate the joules needed to heat 32.1 g of aluminum from 21.0∘C21.0^{\circ} \mathrm{C}21.0∘C to 29.5∘C29.5^{\circ} \mathrm{C}29.5∘C using 0.897 J/g∘C0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}0.897 J/g∘C.

Studdy Solution

STEP 1

STEP 2

First, we need to calculate the change in temperature. 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=29.5∘C−21.0∘C\Delta =29.5^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C}Δ=29.5∘C−21.0∘C

STEP 4

Calculate the change in temperature.Δ=29.∘C−21.0∘C=8.∘C\Delta =29.^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C} =8.^{\circ} \mathrm{C}Δ=29.∘C−21.0∘C=8.∘C

STEP 5

Now that we have the change in temperature, we can calculate the amount of heat required to heat the aluminum. We can do this by using the formula q=mcΔq = mc\Deltaq=mcΔ.q=m×c×Δq = m \times c \times \Deltaq=m×c×Δ

STEP 6

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat.q=32.1 g×0.897 J/g∘C×8.5∘Cq =32.1 \mathrm{~g} \times0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times8.5^{\circ} \mathrm{C}q=32.1 g×0.897 J/g∘C×8.5∘C

STEP 7

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QuestionCalculate the joules needed to heat 32.1 g of aluminum from $21.0^{\circ} \mathrm{C}$ to $29.5^{\circ} \mathrm{C}$ using $0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ .

First, we need to calculate the change in temperature. 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.
$\Delta =29.5^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C}$

Calculate the change in temperature.
$\Delta =29.^{\circ} \mathrm{C} -21.0^{\circ} \mathrm{C} =8.^{\circ} \mathrm{C}$

Now that we have the change in temperature, we can calculate the amount of heat required to heat the aluminum. We can do this by using the formula $q = mc\Delta$ .
$q = m \times c \times \Delta$

Plug in the values for the mass, specific heat, and change in temperature to calculate the heat.
$q =32.1 \mathrm{~g} \times0.897 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times8.5^{\circ} \mathrm{C}$