Math / Algebra

Question2a) Consider a room in a house that has composite walls with an $R_{t o t}=1.0 \mathrm{~m}^{2} \mathrm{~K} / \mathrm{W}$ . The total cross sectional area of the walls is $35 \mathrm{~m}^{2}$ . During winter, the temperature outside the room is $T_{\text {out }}=6^{\circ} \mathrm{C}$ . Inside the room, there is an electric heater that emits heat at a rate of 220 W that keeps the room warm. The room is occupied by 3 people that emit heat at a rate 80 W each. What is the temperature inside the room $T_{i n}$ (in ${ }^{\circ} \mathrm{C}$ ) in winter? Express your answer with 1 decimal place.
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Studdy Solution

STEP 1

What is this asking? We need to find the temperature inside a room heated by an electric heater and some people, given the outside temperature, wall insulation, and heat emission rates. Watch out! Don't forget to convert the total heat generated by the people to watts, and make sure all temperatures are in Celsius.

STEP 2

1. Calculate total heat emitted by people.
2. Calculate total heat input.
3. Calculate the temperature difference.
4. Calculate the inside temperature.

STEP 3

Alright, so we've got 3 people each giving off 80 W of heat.
Let's multiply to find the total heat from the people!

STEP 4

$3 \cdot 80 \text{ W} = 240 \text{ W}$ So, the people are generating a total of 240 W of heat.
Nice!

STEP 5

Now, let's add the heat from the 220 W electric heater to the 240 W from the people to get the total heat input $Q_{in}$ .

STEP 6

$Q_{in} = 220 \text{ W} + 240 \text{ W} = 460 \text{ W}$ Awesome! The total heat input into the room is 460 W.

STEP 7

We know the total thermal resistance $R_{tot}$ is 1.0 $\text{m}^2 \cdot \text{K} / \text{W}$ and the total area $A$ of the walls is 35 $\text{m}^2$ .
We can use these to find the overall heat transfer coefficient $U$ .
Remember, $U$ is just the inverse of $R_{tot}$ multiplied by the area.

STEP 8

$U = \frac{1}{R_{tot}} \cdot A = \frac{1}{1.0 \text{ m}^2 \cdot \text{K} / \text{W}} \cdot 35 \text{ m}^2 = 35 \text{ W/K}$ So, our overall heat transfer coefficient $U$ is 35 W/K.

STEP 9

The heat transfer rate $Q$ is related to the temperature difference $\Delta T$ and the overall heat transfer coefficient $U$ by the equation $Q = U \cdot \Delta T$ .
We know $Q$ (which is the same as $Q_{in}$ from before) and $U$ , so we can solve for $\Delta T$ .

STEP 10

$\Delta T = \frac{Q}{U} = \frac{460 \text{ W}}{35 \text{ W/K}} \approx 13.1 \text{ K}$ This means the temperature difference between the inside and outside is about 13.1 K.
Since Kelvin and Celsius degrees are the same size, this is also a 13.1°C difference.

STEP 11

We know the outside temperature $T_{out}$ is 6°C, and the inside is 13.1°C warmer.
Let's add them up!

STEP 12

$T_{in} = T_{out} + \Delta T = 6^\circ \text{C} + 13.1^\circ \text{C} = 19.1^\circ \text{C}$ So, the inside temperature is 19.1°C.
Perfect!

STEP 13

The temperature inside the room is 19.1°C.

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Studdy Solution

STEP 1

STEP 2

1. Calculate total heat emitted by people.
2. Calculate total heat input.
3. Calculate the temperature difference.
4. Calculate the inside temperature.

STEP 3

Alright, so we've got 3 people each giving off 80 W of heat.
Let's multiply to find the total heat from the people!

STEP 4

$3 \cdot 80 \text{ W} = 240 \text{ W}$ So, the people are generating a total of 240 W of heat.
Nice!

STEP 5

Now, let's add the heat from the 220 W electric heater to the 240 W from the people to get the total heat input $Q_{in}$ .

STEP 6

$Q_{in} = 220 \text{ W} + 240 \text{ W} = 460 \text{ W}$ Awesome! The total heat input into the room is 460 W.

STEP 7

We know the total thermal resistance $R_{tot}$ is 1.0 $\text{m}^2 \cdot \text{K} / \text{W}$ and the total area $A$ of the walls is 35 $\text{m}^2$ .
We can use these to find the overall heat transfer coefficient $U$ .
Remember, $U$ is just the inverse of $R_{tot}$ multiplied by the area.

STEP 8

$U = \frac{1}{R_{tot}} \cdot A = \frac{1}{1.0 \text{ m}^2 \cdot \text{K} / \text{W}} \cdot 35 \text{ m}^2 = 35 \text{ W/K}$ So, our overall heat transfer coefficient $U$ is 35 W/K.

STEP 9

The heat transfer rate $Q$ is related to the temperature difference $\Delta T$ and the overall heat transfer coefficient $U$ by the equation $Q = U \cdot \Delta T$ .
We know $Q$ (which is the same as $Q_{in}$ from before) and $U$ , so we can solve for $\Delta T$ .

STEP 10

$\Delta T = \frac{Q}{U} = \frac{460 \text{ W}}{35 \text{ W/K}} \approx 13.1 \text{ K}$ This means the temperature difference between the inside and outside is about 13.1 K.
Since Kelvin and Celsius degrees are the same size, this is also a 13.1°C difference.

STEP 11

We know the outside temperature $T_{out}$ is 6°C, and the inside is 13.1°C warmer.
Let's add them up!

STEP 12

$T_{in} = T_{out} + \Delta T = 6^\circ \text{C} + 13.1^\circ \text{C} = 19.1^\circ \text{C}$ So, the inside temperature is 19.1°C.
Perfect!

STEP 13

The temperature inside the room is 19.1°C.

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Studdy Solution

STEP 1

STEP 2

1. Calculate total heat emitted by people.2. Calculate total heat input.3. Calculate the temperature difference.4. Calculate the inside temperature.

STEP 3

Alright, so we've got **3** people each giving off **80 W** of heat.Let's multiply to find the total heat from the people!

STEP 4

3⋅80 W=240 W 3 \cdot 80 \text{ W} = 240 \text{ W} 3⋅80 W=240 W So, the people are generating a total of **240 W** of heat.Nice!

STEP 5

Now, let's add the heat from the **220 W** electric heater to the **240 W** from the people to get the *total* heat input QinQ_{in}Qin​.

STEP 6

Qin=220 W+240 W=460 W Q_{in} = 220 \text{ W} + 240 \text{ W} = 460 \text{ W} Qin​=220 W+240 W=460 W Awesome! The total heat input into the room is **460 W**.

STEP 7

STEP 8

U=1Rtot⋅A=11.0 m2⋅K/W⋅35 m2=35 W/K U = \frac{1}{R_{tot}} \cdot A = \frac{1}{1.0 \text{ m}^2 \cdot \text{K} / \text{W}} \cdot 35 \text{ m}^2 = 35 \text{ W/K} U=Rtot​1​⋅A=1.0 m2⋅K/W1​⋅35 m2=35 W/K So, our overall heat transfer coefficient UUU is **35 W/K**.

STEP 9

The heat transfer rate QQQ is related to the temperature difference ΔT\Delta TΔT and the overall heat transfer coefficient UUU by the equation Q=U⋅ΔTQ = U \cdot \Delta TQ=U⋅ΔT.We know QQQ (which is the same as QinQ_{in}Qin​ from before) and UUU, so we can solve for ΔT\Delta TΔT.

STEP 10

STEP 11

We know the outside temperature ToutT_{out}Tout​ is **6°C**, and the inside is **13.1°C** warmer.Let's add them up!

STEP 12

Tin=Tout+ΔT=6∘C+13.1∘C=19.1∘C T_{in} = T_{out} + \Delta T = 6^\circ \text{C} + 13.1^\circ \text{C} = 19.1^\circ \text{C} Tin​=Tout​+ΔT=6∘C+13.1∘C=19.1∘C So, the inside temperature is **19.1°C**.Perfect!

STEP 13

The temperature inside the room is 19.1°C.

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1. Calculate total heat emitted by people.
2. Calculate total heat input.
3. Calculate the temperature difference.
4. Calculate the inside temperature.

Alright, so we've got 3 people each giving off 80 W of heat.
Let's multiply to find the total heat from the people!

$3 \cdot 80 \text{ W} = 240 \text{ W}$ So, the people are generating a total of 240 W of heat.
Nice!

Now, let's add the heat from the 220 W electric heater to the 240 W from the people to get the total heat input $Q_{in}$ .

$Q_{in} = 220 \text{ W} + 240 \text{ W} = 460 \text{ W}$ Awesome! The total heat input into the room is 460 W.

$U = \frac{1}{R_{tot}} \cdot A = \frac{1}{1.0 \text{ m}^2 \cdot \text{K} / \text{W}} \cdot 35 \text{ m}^2 = 35 \text{ W/K}$ So, our overall heat transfer coefficient $U$ is 35 W/K.

The heat transfer rate $Q$ is related to the temperature difference $\Delta T$ and the overall heat transfer coefficient $U$ by the equation $Q = U \cdot \Delta T$ .
We know $Q$ (which is the same as $Q_{in}$ from before) and $U$ , so we can solve for $\Delta T$ .

We know the outside temperature $T_{out}$ is 6°C, and the inside is 13.1°C warmer.
Let's add them up!

$T_{in} = T_{out} + \Delta T = 6^\circ \text{C} + 13.1^\circ \text{C} = 19.1^\circ \text{C}$ So, the inside temperature is 19.1°C.
Perfect!