QuestionCalculate the mass of Titanium (Ti) that absorbs $1630 \, \mathrm{J}$ , with a specific heat of $0.523 \, \mathrm{J/g^{\circ}C}$ , and a temp change from $69 \, ^{\circ} \mathrm{C}$ to $188 \, ^{\circ} \mathrm{C}$ .

Studdy Solution

STEP 1

Assumptions1. The specific heat capacity of Titanium (Ti) is $0.523 \, \mathrm{J/g^{\circ}C}$ . The sample absorbs $1630 \, \mathrm{J}$ of heat3. The temperature of the sample increases from $69 \, ^{\circ} \mathrm{C}$ to $188 \, ^{\circ} \mathrm{C}$
4. The formula for heat absorbed or released by a substance is given by $q = mc\Delta$ , where $q$ is the heat absorbed or released, $m$ is the mass of the substance, $c$ is the specific heat capacity of the substance, and $\Delta$ is the change in temperature.

STEP 2

First, we need to find 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.
$\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C}$

STEP 4

Calculate the change in temperature.
$\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C} =119 \, ^{\circ} \mathrm{C}$

STEP 5

Now that we have the change in temperature, we can find the mass of the Titanium sample. We can rearrange the formula for heat absorbed to solve for mass.
$m = \frac{q}{c\Delta}$

STEP 6

Plug in the values for the heat absorbed, the specific heat capacity, and the change in temperature to calculate the mass.
$m = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}}$

STEP 7

Calculate the mass of the Titanium sample.
$m = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}} =26.1 \, \mathrm{g}$ The mass of the Titanium sample is $26.1 \, \mathrm{g}$ .

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QuestionCalculate the mass of Titanium (Ti) that absorbs $1630 \, \mathrm{J}$ , with a specific heat of $0.523 \, \mathrm{J/g^{\circ}C}$ , and a temp change from $69 \, ^{\circ} \mathrm{C}$ to $188 \, ^{\circ} \mathrm{C}$ .

Studdy Solution

STEP 1

STEP 2

First, we need to find 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.
$\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C}$

STEP 4

Calculate the change in temperature.
$\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C} =119 \, ^{\circ} \mathrm{C}$

STEP 5

Now that we have the change in temperature, we can find the mass of the Titanium sample. We can rearrange the formula for heat absorbed to solve for mass.
$m = \frac{q}{c\Delta}$

STEP 6

Plug in the values for the heat absorbed, the specific heat capacity, and the change in temperature to calculate the mass.
$m = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}}$

STEP 7

Calculate the mass of the Titanium sample.
$m = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}} =26.1 \, \mathrm{g}$ The mass of the Titanium sample is $26.1 \, \mathrm{g}$ .

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QuestionCalculate the mass of Titanium (Ti) that absorbs 1630 J1630 \, \mathrm{J}1630J, with a specific heat of 0.523 J/g∘C0.523 \, \mathrm{J/g^{\circ}C}0.523J/g∘C, and a temp change from 69 ∘C69 \, ^{\circ} \mathrm{C}69∘C to 188 ∘C188 \, ^{\circ} \mathrm{C}188∘C.

Studdy Solution

STEP 1

STEP 2

First, we need to find 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=188 ∘C−69 ∘C\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C}Δ=188∘C−69∘C

STEP 4

Calculate the change in temperature.Δ=188 ∘C−69 ∘C=119 ∘C\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C} =119 \, ^{\circ} \mathrm{C}Δ=188∘C−69∘C=119∘C

STEP 5

Now that we have the change in temperature, we can find the mass of the Titanium sample. We can rearrange the formula for heat absorbed to solve for mass.m=qcΔm = \frac{q}{c\Delta}m=cΔq​

STEP 6

Plug in the values for the heat absorbed, the specific heat capacity, and the change in temperature to calculate the mass.m=1630 J0.523 J/g∘C×119 ∘Cm = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}}m=0.523J/g∘C×119∘C1630J​

STEP 7

Calculate the mass of the Titanium sample.m=1630 J0.523 J/g∘C×119 ∘C=26.1 gm = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}} =26.1 \, \mathrm{g}m=0.523J/g∘C×119∘C1630J​=26.1gThe mass of the Titanium sample is 26.1 g26.1 \, \mathrm{g}26.1g.

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QuestionCalculate the mass of Titanium (Ti) that absorbs $1630 \, \mathrm{J}$ , with a specific heat of $0.523 \, \mathrm{J/g^{\circ}C}$ , and a temp change from $69 \, ^{\circ} \mathrm{C}$ to $188 \, ^{\circ} \mathrm{C}$ .

First, we need to find 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.
$\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C}$

Calculate the change in temperature.
$\Delta =188 \, ^{\circ} \mathrm{C} -69 \, ^{\circ} \mathrm{C} =119 \, ^{\circ} \mathrm{C}$

Now that we have the change in temperature, we can find the mass of the Titanium sample. We can rearrange the formula for heat absorbed to solve for mass.
$m = \frac{q}{c\Delta}$

Plug in the values for the heat absorbed, the specific heat capacity, and the change in temperature to calculate the mass.
$m = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}}$

Calculate the mass of the Titanium sample.
$m = \frac{1630 \, \mathrm{J}}{0.523 \, \mathrm{J/g^{\circ}C} \times119 \, ^{\circ} \mathrm{C}} =26.1 \, \mathrm{g}$ The mass of the Titanium sample is $26.1 \, \mathrm{g}$ .