A 10.0g sample of H2O(l) at 23.0degrees C absorbs 209 joules of heat. What is the final temp of the H2O(l) sample?

Its 28.O C

Gavin is right its 28.0c. NOT what SCOTT said he is WRONG.

My answer isn't wrong. a 5ºC increase is also equal to 28ºC. 23 + 5 = 28. Sorry if it was misinterpreted.

this is a tutor website, not cheat site

If this is a so-called "cheating" website, why are you on it? Smh, leave us alone, and go worry about yourself.

no

To find the final temperature of the H2O(l) sample, we need to use the heat capacity formula.

The heat capacity formula is given by Q = m × C × ΔT, where:
- Q is the heat absorbed or released by the substance,
- m is the mass of the substance,
- C is the specific heat capacity of the substance, and
- ΔT is the change in temperature.

In this case, we are given:
- Q = 209 joules,
- m = 10.0g,
- C = specific heat capacity of water (4.18 J/g°C - this value is commonly known and can be found in reference materials), and
- ΔT is the change in temperature.

We can rearrange the formula to find ΔT:
ΔT = Q / (m × C)

Substituting the given values, we have:
ΔT = 209 J / (10.0g × 4.18 J/g°C)

Calculating this, we get:
ΔT ≈ 5.0°C

To find the final temperature, we add the change in temperature to the initial temperature of 23.0°C:
Final temperature = Initial temperature + ΔT
Final temperature ≈ 23.0°C + 5.0°C

Therefore, the final temperature of the H2O(l) sample is approximately 28.0°C.

Thank you so much Scott ! .

it takes 4.179 J of heat to change the temperature of 1 g of water by one ºC

209 / (10.0 * 4.179) = temp. change

looks like about a 5ºC increase

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