Please can any one help me by answering this question!!+ignore my answer!,

A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I

murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

Step 1: data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

Step 2: Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

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3 years ago

Which stament about energy is mostly correct

klasskru [66]

I would say C is the most correct.
In D it depends on what water source you're using. Let's say it is a waterfall, then the source of the water (melting ice or a lake) may disappear in the future.
If you're using underwater "windmills" placed in the ocean, then you would expect it to last a while as the ocean will not disappear in the near future.

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3 years ago

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Firlakuza [10]

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cricket20 [7]

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3 years ago

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Yuki888 [10]

Answer:

True