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Agata [3.3K]
3 years ago
13

You and your lab group have been asked to design an investigation to determine the effects of heat transfer between two differen

t objects.
Which of the following investigation designs explains the heat transfer when atoms directly touch?
А
Boiling water, in a pan, sitting on a lit Bunsen burner.
B
Boiling water, in a pan, in front of a lit Bunsen burner.
С
Hands warming in front of a lit Bunsen burner.
D
Hands touching the handle of pan sitting on a lit Bunsen burner.
Chemistry
1 answer:
GREYUIT [131]3 years ago
5 0
I believe that answer is D
The heat from the Bunsen burner transfers to the water and the pot, then the heat from the pot transfers to the person’s hand.
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14. In the lab, an experimenter mixes 75.0 g of water (initially at 30oC) with 83.8 g of a solid metal (initially at 600oC). At
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Answer:

Tungsten is used for this experiment

Explanation:

This is a Thermal - equilibrium situation. we can use the equation.

Loss of Heat of the Metal = Gain of Heat by the Water

                      -Q_{m}=+Q_{w}\\

                    Q = mΔTC_{p}

Q = heat

m = mass

ΔT = T₂ - T₁

T₂ = final temperature

T₁ = Initial temperature

Cp = Specific heat capacity

<u>Metal</u>

m = 83.8 g

T₂ = 50⁰C

T₁ = 600⁰C

Cp = x

<u>Water</u>

m = 75 g

T₂ = 50⁰C

T₁ = 30⁰C

Cp = 4.184 j.g⁻¹.⁰c⁻¹

               -Q_{m}=+Q_{w}\\

⇒ - 83.8 x x x (50 - 600) = 75 x 4.184 x (50 - 30)

⇒ x = \frac{6276}{46090} = 0.13 j.g⁻¹.⁰c⁻¹

We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹

So metal Tungsten used in this experiment

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A serving of Cheez-Its releases 1.30 x 10^4 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy we
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Answer:

The final temperature is:- 7428571463.57 °C

Explanation:

The expression for the calculation of heat is shown below as:-

Q=m\times C\times \Delta T

Where,  

Q  is the heat absorbed/released

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)

Specific heat of water = 4.18 J/g°C

Initial temperature = 35 °C

Final temperature = x °C

\Delta T=(x-35)\ ^0C/tex]&#10;Q = [tex]1.3\times 10^4 kcal

Also, 1 kcal = 4.18 kJ = 4.18\times 10^3 J

So, Q = 1.3\times 10^4\times 4.18\times 10^3 J = 54340000 J

So,  

54340000=0.00175\times 4.18\times (x-35)

0.00175\times \:4.18\left(x-35\right)=54340000

x-35=\frac{54340000}{0.007315}

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