Math / Algebra

QuestionCalculate the energy required to heat 295.0 mg of ammonia from $24.1^{\circ} \mathrm{C}$ to $44.0^{\circ} \mathrm{C}$ . Assume the specific heat capacity of ammonia under these conditions is $4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}$ . Round your answer to 3 significant digits. $\square$

Studdy Solution

STEP 1

What is this asking? How much energy do we need to make 295 milligrams of ammonia a bit hotter? Watch out! Don't forget to convert milligrams to grams and Celsius to Kelvin!
Also, keep track of those units!

STEP 2

1. Convert units
2. Calculate the change in temperature
3. Calculate the energy

STEP 3

Alright, let's convert those milligrams of ammonia to grams!
We know that there are $\text{1000 mg}$ in $\text{1 g}$ .
So, we're going to divide our $\text{295.0 mg}$ by $\text{1000 mg/g}$ to get the mass in grams.
It's like figuring out how many groups of 1000 milligrams fit into 295 milligrams.

STEP 4

$\frac{\text{295.0 mg}}{1} \cdot \frac{\text{1 g}}{\text{1000 mg}} = \text{0.295 g}$ So, we have 0.295 g of ammonia.
Perfect!

STEP 5

Now, let's find the change in temperature.
We're heating the ammonia from $24.1^\circ \text{C}$ to $44.0^\circ \text{C}$ .
The change in temperature in Celsius is the same as the change in temperature in Kelvin, so we can just subtract the initial temperature from the final temperature in Celsius.

STEP 6

$44.0^\circ \text{C} - 24.1^\circ \text{C} = 19.9^\circ \text{C}$ Which is also a change of $\text{19.9 K}$ .
So our change in temperature is $\text{19.9 K}$ .

STEP 7

Time to use our handy-dandy formula for calculating the energy change: $q = mc\Delta T$ Where $q$ is the heat absorbed or released, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

STEP 8

We know the mass ( $m = \text{0.295 g}$ ), the specific heat capacity ( $c = 4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}$ ), and the change in temperature ( $\Delta T = \text{19.9 K}$ ).
Let's plug those values into our formula!

STEP 9

$q = (\text{0.295 g}) \cdot (4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}) \cdot (\text{19.9 K})$

STEP 10

Notice how the units of grams ( $\text{g}$ ) and Kelvin ( $\text{K}$ ) will divide to one, leaving us with Joules ( $\text{J}$ ), which is exactly what we want for energy!

STEP 11

$q = \text{27.71155 J}$ Rounding to 3 significant digits, we get: $q = \text{27.7 J}$

STEP 12

It takes 27.7 J of energy to heat the ammonia.

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Studdy Solution

STEP 1

What is this asking? How much energy do we need to make 295 milligrams of ammonia a bit hotter? Watch out! Don't forget to convert milligrams to grams and Celsius to Kelvin!
Also, keep track of those units!

STEP 2

1. Convert units
2. Calculate the change in temperature
3. Calculate the energy

STEP 3

Alright, let's convert those milligrams of ammonia to grams!
We know that there are $\text{1000 mg}$ in $\text{1 g}$ .
So, we're going to divide our $\text{295.0 mg}$ by $\text{1000 mg/g}$ to get the mass in grams.
It's like figuring out how many groups of 1000 milligrams fit into 295 milligrams.

STEP 4

$\frac{\text{295.0 mg}}{1} \cdot \frac{\text{1 g}}{\text{1000 mg}} = \text{0.295 g}$ So, we have 0.295 g of ammonia.
Perfect!

STEP 5

Now, let's find the change in temperature.
We're heating the ammonia from $24.1^\circ \text{C}$ to $44.0^\circ \text{C}$ .
The change in temperature in Celsius is the same as the change in temperature in Kelvin, so we can just subtract the initial temperature from the final temperature in Celsius.

STEP 6

$44.0^\circ \text{C} - 24.1^\circ \text{C} = 19.9^\circ \text{C}$ Which is also a change of $\text{19.9 K}$ .
So our change in temperature is $\text{19.9 K}$ .

STEP 7

Time to use our handy-dandy formula for calculating the energy change: $q = mc\Delta T$ Where $q$ is the heat absorbed or released, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

STEP 8

We know the mass ( $m = \text{0.295 g}$ ), the specific heat capacity ( $c = 4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}$ ), and the change in temperature ( $\Delta T = \text{19.9 K}$ ).
Let's plug those values into our formula!

STEP 9

$q = (\text{0.295 g}) \cdot (4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}) \cdot (\text{19.9 K})$

STEP 10

Notice how the units of grams ( $\text{g}$ ) and Kelvin ( $\text{K}$ ) will divide to one, leaving us with Joules ( $\text{J}$ ), which is exactly what we want for energy!

STEP 11

$q = \text{27.71155 J}$ Rounding to 3 significant digits, we get: $q = \text{27.7 J}$

STEP 12

It takes 27.7 J of energy to heat the ammonia.

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Studdy Solution

STEP 1

What is this asking? How much energy do we need to make 295 milligrams of ammonia a bit hotter? Watch out! Don't forget to convert milligrams to grams and Celsius to Kelvin!Also, keep track of those units!

STEP 2

1. Convert units2. Calculate the change in temperature3. Calculate the energy

STEP 3

STEP 4

295.0 mg1⋅1 g1000 mg=0.295 g \frac{\text{295.0 mg}}{1} \cdot \frac{\text{1 g}}{\text{1000 mg}} = \text{0.295 g} 1295.0 mg​⋅1000 mg1 g​=0.295 g So, we have **0.295 g** of ammonia.Perfect!

STEP 5

STEP 6

44.0∘C−24.1∘C=19.9∘C 44.0^\circ \text{C} - 24.1^\circ \text{C} = 19.9^\circ \text{C} 44.0∘C−24.1∘C=19.9∘C Which is also a change of 19.9 K\text{19.9 K}19.9 K.So our **change in temperature** is 19.9 K\text{19.9 K}19.9 K.

STEP 7

Time to use our handy-dandy formula for **calculating the energy** change: q=mcΔT q = mc\Delta T q=mcΔT Where qqq is the **heat absorbed or released**, mmm is the **mass**, ccc is the **specific heat capacity**, and ΔT\Delta TΔT is the **change in temperature**.

STEP 8

STEP 9

q=(0.295 g)⋅(4.70 J⋅ g−1⋅ K−1)⋅(19.9 K) q = (\text{0.295 g}) \cdot (4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}) \cdot (\text{19.9 K}) q=(0.295 g)⋅(4.70 J⋅ g−1⋅ K−1)⋅(19.9 K)

STEP 10

Notice how the units of grams (g\text{g}g) and Kelvin (K\text{K}K) will **divide to one**, leaving us with Joules (J\text{J}J), which is exactly what we want for energy!

STEP 11

q=27.71155 J q = \text{27.71155 J} q=27.71155 J Rounding to **3 significant digits**, we get: q=27.7 J q = \text{27.7 J} q=27.7 J

STEP 12

It takes **27.7 J** of energy to heat the ammonia.

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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.

What is this asking? How much energy do we need to make 295 milligrams of ammonia a bit hotter? Watch out! Don't forget to convert milligrams to grams and Celsius to Kelvin!
Also, keep track of those units!

1. Convert units
2. Calculate the change in temperature
3. Calculate the energy

$\frac{\text{295.0 mg}}{1} \cdot \frac{\text{1 g}}{\text{1000 mg}} = \text{0.295 g}$ So, we have 0.295 g of ammonia.
Perfect!

$44.0^\circ \text{C} - 24.1^\circ \text{C} = 19.9^\circ \text{C}$ Which is also a change of $\text{19.9 K}$ .
So our change in temperature is $\text{19.9 K}$ .

Time to use our handy-dandy formula for calculating the energy change: $q = mc\Delta T$ Where $q$ is the heat absorbed or released, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

$q = (\text{0.295 g}) \cdot (4.70 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}) \cdot (\text{19.9 K})$

Notice how the units of grams ( $\text{g}$ ) and Kelvin ( $\text{K}$ ) will divide to one, leaving us with Joules ( $\text{J}$ ), which is exactly what we want for energy!

$q = \text{27.71155 J}$ Rounding to 3 significant digits, we get: $q = \text{27.7 J}$

It takes 27.7 J of energy to heat the ammonia.