Imagine a scientific scene depicting thermal energy transfer. Visualize a beaker with 100g of water which has an initial temperature of 20.0°C. Illustrate the concept of this water absorbing 1000 J of thermal energy, perhaps by using abstract forms or symbolic representations for energy – like glowing particles or symbols traditionally used in physics. Avoid showing any calculations or textual information, focusing only on concepts: water, initial temperature, and energy absorption. Please, do not include any text in the image.

Determine the final temperature when 100 g of water absorbs 1000 J of thermal energy, given that the initial temperature of the water is 20.0°C. The theoretical specific heat capacity of water is 4.19 J/g·°C. Show your work below.

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The final temperature of the water can be determined using the formula:

Q = mcΔT

where Q is the thermal energy absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Substituting the given values, we have:

1000 J = (100 g) x (4.19 J/g·°C) x ΔT

Simplifying, we get:

ΔT = 1000 J / (100 g x 4.19 J/g·°C)

ΔT = 2.391°C

The change in temperature is positive, indicating that the water has gained heat. To find the final temperature, we add the change in temperature to the initial temperature:

Final temperature = 20.0°C + 2.391°C

Final temperature = 22.391°C

Therefore, the final temperature of the water is 22.391°C (rounded to three decimal places).

Well, according to my calculations... or should I say, my clown-ulations! We can use the formula Q = mcΔT, where Q is the thermal energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.

So, let's plug in the given values:
Q = 1000 J
m = 100 g
c = 4.19 J/g·°C
ΔT = ?

To find ΔT, we rearrange the formula and solve for it:
ΔT = Q / (mc)
ΔT = 1000 J / (100 g * 4.19 J/g·°C)
ΔT ≈ 2.396 °C

Now we just need to add ΔT to the initial temperature to find the final temperature:
Final temperature = 20.0°C + 2.396°C
Final temperature ≈ 22.396°C

So, the final temperature when 100 g of water absorbs 1000 J of thermal energy is approximately 22.396°C. Keep in mind, my calculations may be clown-ish, but hey, at least I'm trying to make thermodynamics funny!

To determine the final temperature when 100 g of water absorbs 1000 J of thermal energy, we can use the formula:

Q = m * c * ΔT

where:
Q is the thermal energy absorbed (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·°C), and
ΔT is the change in temperature (in °C).

In this case, we are given:
Q = 1000 J
m = 100 g
c = 4.19 J/g·°C
Initial temperature (T₁) = 20.0°C

We want to find the final temperature (T₂).

First, let's rearrange the formula to solve for ΔT:

ΔT = Q / (m * c)

Plugging in the values, we have:

ΔT = 1000 J / (100 g * 4.19 J/g·°C)
ΔT ≈ 2.39 °C

Now, we can find the final temperature (T₂) by adding the change in temperature to the initial temperature:

T₂ = T₁ + ΔT
T₂ = 20.0°C + 2.39°C
T₂ ≈ 22.39°C

Therefore, the final temperature of the water is approximately 22.39°C.