QuestionA water sample absorbed $2,851 \mathrm{~J}$ from a $42.6 \mathrm{~g}$ metal piece cooling from $209^{\circ} \mathrm{C}$ to $40.0^{\circ} \mathrm{C}$ . Find the specific heat of the metal: $\mathrm{C}_{\text {metal }}=[?] \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

Studdy Solution

STEP 1

Assumptions1. The energy absorbed by the water sample is,851 J. The mass of the metal piece is42.6 g3. The initial temperature of the metal is209°C4. The final temperature of the metal is40.0°C5. The specific heat of the metal is unknown and is what we are trying to find.

STEP 2

The formula for heat transfer is given by $q = m \cdot C \cdot \Delta$ where- $q$ is the heat energy 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

We can rearrange the formula to solve for the specific heat capacity $C$ $C = \frac{q}{m \cdot \Delta}$

STEP 4

The change in temperature $\Delta$ is given by the final temperature minus the initial temperature. In this case, the metal is cooling down, so $\Delta$ will be negative.
$\Delta =_{final} -_{initial}$

STEP 5

Substitute the given values into the equation for $\Delta$ $\Delta =40.0°C -209°C$

STEP 6

Calculate the change in temperature $\Delta =40.0°C -209°C = -169°C$

STEP 7

Now, substitute the given values into the equation for $C$ $C = \frac{2851 J}{42.6 g \cdot -169°C}$

STEP 8

Calculate the specific heat capacity of the metal $C = \frac{2851 J}{42.6 g \cdot -169°C} = -0.399 \frac{J}{g \cdot °C}$ The specific heat capacity cannot be negative, so we take the absolute value to get the final answer.

STEP 9

Take the absolute value of $C$ $C = |-.399 \frac{J}{g \cdot °C}| =.399 \frac{J}{g \cdot °C}$ The specific heat capacity of the metal is $.399 \frac{J}{g \cdot °C}$ .

Was this helpful?

QuestionA water sample absorbed $2,851 \mathrm{~J}$ from a $42.6 \mathrm{~g}$ metal piece cooling from $209^{\circ} \mathrm{C}$ to $40.0^{\circ} \mathrm{C}$ . Find the specific heat of the metal: $\mathrm{C}_{\text {metal }}=[?] \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}$ .

Studdy Solution

STEP 1

Assumptions1. The energy absorbed by the water sample is,851 J. The mass of the metal piece is42.6 g3. The initial temperature of the metal is209°C4. The final temperature of the metal is40.0°C5. The specific heat of the metal is unknown and is what we are trying to find.

STEP 2

The formula for heat transfer is given by $q = m \cdot C \cdot \Delta$ where- $q$ is the heat energy 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

We can rearrange the formula to solve for the specific heat capacity $C$ $C = \frac{q}{m \cdot \Delta}$

STEP 4

The change in temperature $\Delta$ is given by the final temperature minus the initial temperature. In this case, the metal is cooling down, so $\Delta$ will be negative.
$\Delta =_{final} -_{initial}$

STEP 5

Substitute the given values into the equation for $\Delta$ $\Delta =40.0°C -209°C$

STEP 6

Calculate the change in temperature $\Delta =40.0°C -209°C = -169°C$

STEP 7

Now, substitute the given values into the equation for $C$ $C = \frac{2851 J}{42.6 g \cdot -169°C}$

STEP 8

Calculate the specific heat capacity of the metal $C = \frac{2851 J}{42.6 g \cdot -169°C} = -0.399 \frac{J}{g \cdot °C}$ The specific heat capacity cannot be negative, so we take the absolute value to get the final answer.

STEP 9

Take the absolute value of $C$ $C = |-.399 \frac{J}{g \cdot °C}| =.399 \frac{J}{g \cdot °C}$ The specific heat capacity of the metal is $.399 \frac{J}{g \cdot °C}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

Assumptions1. The energy absorbed by the water sample is,851 J. The mass of the metal piece is42.6 g3. The initial temperature of the metal is209°C4. The final temperature of the metal is40.0°C5. The specific heat of the metal is unknown and is what we are trying to find.

STEP 2

The formula for heat transfer is given byq=m⋅C⋅Δq = m \cdot C \cdot \Deltaq=m⋅C⋅Δwhere- qqq is the heat energy 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

We can rearrange the formula to solve for the specific heat capacity CCCC=qm⋅ΔC = \frac{q}{m \cdot \Delta}C=m⋅Δq​

STEP 4

The change in temperature Δ\DeltaΔ is given by the final temperature minus the initial temperature. In this case, the metal is cooling down, so Δ\DeltaΔ will be negative.Δ=final−initial\Delta =_{final} -_{initial}Δ=final​−initial​

STEP 5

Substitute the given values into the equation for Δ\DeltaΔΔ=40.0°C−209°C\Delta =40.0°C -209°CΔ=40.0°C−209°C

STEP 6

Calculate the change in temperatureΔ=40.0°C−209°C=−169°C\Delta =40.0°C -209°C = -169°CΔ=40.0°C−209°C=−169°C

STEP 7

Now, substitute the given values into the equation for CCCC=2851J42.6g⋅−169°CC = \frac{2851 J}{42.6 g \cdot -169°C}C=42.6g⋅−169°C2851J​

STEP 8

STEP 9

Take the absolute value of CCCC=∣−.399Jg⋅°C∣=.399Jg⋅°CC = |-.399 \frac{J}{g \cdot °C}| =.399 \frac{J}{g \cdot °C}C=∣−.399g⋅°CJ​∣=.399g⋅°CJ​The specific heat capacity of the metal is .399Jg⋅°C.399 \frac{J}{g \cdot °C}.399g⋅°CJ​.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

The formula for heat transfer is given by $q = m \cdot C \cdot \Delta$ where- $q$ is the heat energy 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.

We can rearrange the formula to solve for the specific heat capacity $C$ $C = \frac{q}{m \cdot \Delta}$

The change in temperature $\Delta$ is given by the final temperature minus the initial temperature. In this case, the metal is cooling down, so $\Delta$ will be negative.
$\Delta =_{final} -_{initial}$

Substitute the given values into the equation for $\Delta$ $\Delta =40.0°C -209°C$

Calculate the change in temperature $\Delta =40.0°C -209°C = -169°C$

Now, substitute the given values into the equation for $C$ $C = \frac{2851 J}{42.6 g \cdot -169°C}$

Take the absolute value of $C$ $C = |-.399 \frac{J}{g \cdot °C}| =.399 \frac{J}{g \cdot °C}$ The specific heat capacity of the metal is $.399 \frac{J}{g \cdot °C}$ .