QuestionHow much energy is absorbed by $10.0 \mathrm{~g}$ of nickel when heated from $2.0^{\circ} \mathrm{C}$ to $63.4^{\circ} \mathrm{C}$ ? Use $\mathrm{C}_{\mathrm{Ni}}=0.120 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

Studdy Solution

STEP 1

Assumptions1. The mass of nickel is10.0 g. The initial temperature of the nickel is.0°C3. The final temperature of the water and nickel is63.4°C4. The specific heat capacity of nickel is0.120 J/g°C5. We are asked to find the heat absorbed by the nickel, denoted as q_Ni

STEP 2

The formula to calculate the 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 3

The change in temperature, $\Delta$ , is calculated as the final temperature minus the initial temperature.
$\Delta =_{final} -_{initial}$

STEP 4

Now, plug in the given values for the final and initial temperatures to calculate the change in temperature.
$\Delta =63.4°C -2.0°C$

STEP 5

Calculate the change in temperature.
$\Delta =63.4°C -2.0°C =61.4°C$

STEP 6

Now that we have the change in temperature, we can substitute all the known values into the heat formula.
$q_{Ni} = m_{Ni}c_{Ni}\Delta_{Ni}$

STEP 7

Plug in the values for the mass of nickel, the specific heat capacity of nickel, and the change in temperature to calculate the heat absorbed by the nickel.
$q_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C$

STEP 8

Calculate the heat absorbed by the nickel.
$q_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C =73.68 J$ The nickel absorbed73.68 Joules of heat.

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QuestionHow much energy is absorbed by $10.0 \mathrm{~g}$ of nickel when heated from $2.0^{\circ} \mathrm{C}$ to $63.4^{\circ} \mathrm{C}$ ? Use $\mathrm{C}_{\mathrm{Ni}}=0.120 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

Studdy Solution

STEP 1

Assumptions1. The mass of nickel is10.0 g. The initial temperature of the nickel is.0°C3. The final temperature of the water and nickel is63.4°C4. The specific heat capacity of nickel is0.120 J/g°C5. We are asked to find the heat absorbed by the nickel, denoted as q_Ni

STEP 2

The formula to calculate the 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 3

The change in temperature, $\Delta$ , is calculated as the final temperature minus the initial temperature.
$\Delta =_{final} -_{initial}$

STEP 4

Now, plug in the given values for the final and initial temperatures to calculate the change in temperature.
$\Delta =63.4°C -2.0°C$

STEP 5

Calculate the change in temperature.
$\Delta =63.4°C -2.0°C =61.4°C$

STEP 6

Now that we have the change in temperature, we can substitute all the known values into the heat formula.
$q_{Ni} = m_{Ni}c_{Ni}\Delta_{Ni}$

STEP 7

Plug in the values for the mass of nickel, the specific heat capacity of nickel, and the change in temperature to calculate the heat absorbed by the nickel.
$q_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C$

STEP 8

Calculate the heat absorbed by the nickel.
$q_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C =73.68 J$ The nickel absorbed73.68 Joules of heat.

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Studdy Solution

STEP 1

Assumptions1. The mass of nickel is10.0 g. The initial temperature of the nickel is.0°C3. The final temperature of the water and nickel is63.4°C4. The specific heat capacity of nickel is0.120 J/g°C5. We are asked to find the heat absorbed by the nickel, denoted as q_Ni

STEP 2

The formula to calculate the heat absorbed or released by a substance is given byq=mcΔq = mc\Deltaq=mcΔwhere- qqq is the heat absorbed or released, - mmm is the mass of the substance, - ccc is the specific heat capacity of the substance, and- Δ\DeltaΔ is the change in temperature.

STEP 3

The change in temperature, Δ\DeltaΔ, is calculated as the final temperature minus the initial temperature.Δ=final−initial\Delta =_{final} -_{initial}Δ=final​−initial​

STEP 4

Now, plug in the given values for the final and initial temperatures to calculate the change in temperature.Δ=63.4°C−2.0°C\Delta =63.4°C -2.0°CΔ=63.4°C−2.0°C

STEP 5

Calculate the change in temperature.Δ=63.4°C−2.0°C=61.4°C\Delta =63.4°C -2.0°C =61.4°CΔ=63.4°C−2.0°C=61.4°C

STEP 6

Now that we have the change in temperature, we can substitute all the known values into the heat formula.qNi=mNicNiΔNiq_{Ni} = m_{Ni}c_{Ni}\Delta_{Ni}qNi​=mNi​cNi​ΔNi​

STEP 7

Plug in the values for the mass of nickel, the specific heat capacity of nickel, and the change in temperature to calculate the heat absorbed by the nickel.qNi=10.0g×0.120Jg°C×61.4°Cq_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°CqNi​=10.0g×0.120g°CJ​×61.4°C

STEP 8

Calculate the heat absorbed by the nickel.qNi=10.0g×0.120Jg°C×61.4°C=73.68Jq_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C =73.68 JqNi​=10.0g×0.120g°CJ​×61.4°C=73.68JThe nickel absorbed73.68 Joules of heat.

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The formula to calculate the 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.

The change in temperature, $\Delta$ , is calculated as the final temperature minus the initial temperature.
$\Delta =_{final} -_{initial}$

Now, plug in the given values for the final and initial temperatures to calculate the change in temperature.
$\Delta =63.4°C -2.0°C$

Calculate the change in temperature.
$\Delta =63.4°C -2.0°C =61.4°C$

Now that we have the change in temperature, we can substitute all the known values into the heat formula.
$q_{Ni} = m_{Ni}c_{Ni}\Delta_{Ni}$

Plug in the values for the mass of nickel, the specific heat capacity of nickel, and the change in temperature to calculate the heat absorbed by the nickel.
$q_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C$

Calculate the heat absorbed by the nickel.
$q_{Ni} =10.0g \times0.120 \frac{J}{g°C} \times61.4°C =73.68 J$ The nickel absorbed73.68 Joules of heat.