Math / Numbers & Operations

QuestionCalculate the energy required to heat 722.0 g of water from $37.1^{\circ} \mathrm{C}$ to $58.3^{\circ} \mathrm{C}$ . Assume the specific heat capacity of water under these conditions is $4.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}$ . Round your answer to 3 significant digits.

Studdy Solution

STEP 1

What is this asking? How much energy do we need to make 722 grams of water a bit hotter? Watch out! Don't forget to convert the temperature difference to Kelvin, even though it's given in Celsius!
Also, keep track of your units throughout the calculation.

STEP 2

1. Temperature Change
2. Energy Calculation

STEP 3

Alright, let's start by finding how much hotter we're making the water.
We're going from an initial temperature of $37.1^\circ \mathrm{C}$ to a final temperature of $58.3^\circ \mathrm{C}$ .

STEP 4

The change in temperature is just the final temperature minus the initial temperature: $58.3^\circ \mathrm{C} - 37.1^\circ \mathrm{C} = 21.2^\circ \mathrm{C}$

STEP 5

Now, even though the specific heat capacity is given in units of Kelvin, a change of $1^\circ \mathrm{C}$ is the same as a change of $1 \mathrm{K}$ .
So, our temperature change in Kelvin is: $21.2^\circ \mathrm{C} = 21.2 \mathrm{K}$ See? Not so bad!

STEP 6

Now for the main event!
We'll use the formula that connects energy, mass, specific heat capacity, and change in temperature: $Q = m \cdot c \cdot \Delta T$ Where: $Q$ is the energy we're looking for, $m$ is the mass of the water ( $722.0 \mathrm{~g}$ ), $c$ is the specific heat capacity ( $4.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1}$ ), and $\Delta T$ is the change in temperature ( $21.2 \mathrm{K}$ ).

STEP 7

Let's plug in our values: $Q = 722.0 \mathrm{~g} \cdot 4.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1} \cdot 21.2 \mathrm{K}$

STEP 8

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

STEP 9

Calculate the result: $Q = 64284.672 \mathrm{~J}$

STEP 10

Finally, let's round our answer to 3 significant digits: $Q \approx 64300 \mathrm{~J}$

STEP 11

The energy required to heat the water is approximately $64300 \mathrm{~J}$ .

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Studdy Solution

STEP 1

What is this asking? How much energy do we need to make 722 grams of water a bit hotter? Watch out! Don't forget to convert the temperature difference to Kelvin, even though it's given in Celsius!
Also, keep track of your units throughout the calculation.

STEP 2

1. Temperature Change
2. Energy Calculation

STEP 3

Alright, let's start by finding how much hotter we're making the water.
We're going from an initial temperature of $37.1^\circ \mathrm{C}$ to a final temperature of $58.3^\circ \mathrm{C}$ .

STEP 4

The change in temperature is just the final temperature minus the initial temperature: $58.3^\circ \mathrm{C} - 37.1^\circ \mathrm{C} = 21.2^\circ \mathrm{C}$

STEP 5

Now, even though the specific heat capacity is given in units of Kelvin, a change of $1^\circ \mathrm{C}$ is the same as a change of $1 \mathrm{K}$ .
So, our temperature change in Kelvin is: $21.2^\circ \mathrm{C} = 21.2 \mathrm{K}$ See? Not so bad!

STEP 6

STEP 7

Let's plug in our values: $Q = 722.0 \mathrm{~g} \cdot 4.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1} \cdot 21.2 \mathrm{K}$

STEP 8

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

STEP 9

Calculate the result: $Q = 64284.672 \mathrm{~J}$

STEP 10

Finally, let's round our answer to 3 significant digits: $Q \approx 64300 \mathrm{~J}$

STEP 11

The energy required to heat the water is approximately $64300 \mathrm{~J}$ .

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Studdy Solution

STEP 1

What is this asking? How much energy do we need to make 722 grams of water a bit hotter? Watch out! Don't forget to convert the temperature difference to Kelvin, even though it's given in Celsius!Also, keep track of your units throughout the calculation.

STEP 2

1. Temperature Change2. Energy Calculation

STEP 3

Alright, let's **start** by finding how much hotter we're making the water.We're going from an **initial temperature** of 37.1∘C37.1^\circ \mathrm{C}37.1∘C to a **final temperature** of 58.3∘C58.3^\circ \mathrm{C}58.3∘C.

STEP 4

The **change in temperature** is just the **final temperature** minus the **initial temperature**: 58.3∘C−37.1∘C=21.2∘C58.3^\circ \mathrm{C} - 37.1^\circ \mathrm{C} = 21.2^\circ \mathrm{C}58.3∘C−37.1∘C=21.2∘C

STEP 5

Now, even though the specific heat capacity is given in units of Kelvin, a *change* of 1∘C1^\circ \mathrm{C}1∘C is the same as a *change* of 1K1 \mathrm{K}1K.So, our **temperature change** in Kelvin is: 21.2∘C=21.2K21.2^\circ \mathrm{C} = 21.2 \mathrm{K}21.2∘C=21.2K See? Not so bad!

STEP 6

STEP 7

Let's **plug in** our values: Q=722.0 g⋅4.18 J⋅ g−1⋅ K−1⋅21.2KQ = 722.0 \mathrm{~g} \cdot 4.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1} \cdot 21.2 \mathrm{K}Q=722.0 g⋅4.18 J⋅ g−1⋅ K−1⋅21.2K

STEP 8

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

STEP 9

**Calculate the result**: Q=64284.672 JQ = 64284.672 \mathrm{~J}Q=64284.672 J

STEP 10

Finally, let's **round** our answer to **3 significant digits**: Q≈64300 JQ \approx 64300 \mathrm{~J}Q≈64300 J

STEP 11

The energy required to heat the water is approximately 64300 J64300 \mathrm{~J}64300 J.

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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 722 grams of water a bit hotter? Watch out! Don't forget to convert the temperature difference to Kelvin, even though it's given in Celsius!
Also, keep track of your units throughout the calculation.

1. Temperature Change
2. Energy Calculation

Alright, let's start by finding how much hotter we're making the water.
We're going from an initial temperature of $37.1^\circ \mathrm{C}$ to a final temperature of $58.3^\circ \mathrm{C}$ .

The change in temperature is just the final temperature minus the initial temperature: $58.3^\circ \mathrm{C} - 37.1^\circ \mathrm{C} = 21.2^\circ \mathrm{C}$

Now, even though the specific heat capacity is given in units of Kelvin, a change of $1^\circ \mathrm{C}$ is the same as a change of $1 \mathrm{K}$ .
So, our temperature change in Kelvin is: $21.2^\circ \mathrm{C} = 21.2 \mathrm{K}$ See? Not so bad!

Let's plug in our values: $Q = 722.0 \mathrm{~g} \cdot 4.18 \mathrm{~J} \cdot \mathrm{~g}^{-1} \cdot \mathrm{~K}^{-1} \cdot 21.2 \mathrm{K}$

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

Calculate the result: $Q = 64284.672 \mathrm{~J}$

Finally, let's round our answer to 3 significant digits: $Q \approx 64300 \mathrm{~J}$

The energy required to heat the water is approximately $64300 \mathrm{~J}$ .