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Pie
3 years ago
10

150.0 grams of iron at 95.0 °C, is placed in an insulated container containing 500.0 grams of water. The temperature of the wate

r increases to 27.2°C. What was the initial temperature of the water? The specific heat of water is 4.18 J/g °C and the specific heat of iron is 0.444 J/g °C
Chemistry
1 answer:
Lorico [155]3 years ago
6 0

Explanation:

in it I mass=150g,initial temp 1=95,mass2=500g,temp2=?,final temperature=27.2 C1=0.444C2=4.18

using formula

m1c1(final temp-initial temp1) =m2c2(temp2-final temp)

150x0.444(27.2-95)=500x4.18(?-27.2)

66(-67.8)=2000(?-27.5)

-4474.8=2000?-55000

collect like terms

-4474.8+55000=2000?

50525=2000?

divide both sides by 2000

2000?/2000=50525/2000

initial temperature =25.26 degree Celsius

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Antarctica, almost completely covered in ice, has an area
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<u>Answer:</u> The mass of ice is 2.39\times 10^{22}g

<u>Explanation:</u>

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Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = 182.88\times 10^3cm     (Conversion factor:  1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

We are given:

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Volume of ice = Area × Height of ice = 142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3

Putting values in above equation, we get:

0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g

Hence, the mass of ice is 2.39\times 10^{22}g

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3 years ago
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