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2 years ago

A metal pellet with a mass of 100.0 g, originally at 116°C, is dropped into a cup of water, initially at

Chemistry

1 answer:

Alina [70]2 years ago

6 0

Answer:

C, 42g

Explanation:

In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).

Assuming no heat is lost to surroundings,

the energy lost from metal pellet = energy gained for water

Since E = mc∆T

(energy = mass x specific heat capacity x temperature change)

mc∆T (metal pellet) = mc∆T (water)

100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)

3958.96 = 94.14m

m = 42g

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If 45.0 mL of ethanol (density =0.789g/mol) initially at 6.0°C mix with 45.0 mL of water (density =1.0 g/mol) initially at 28.0°

Likurg_2 [28]

The final temperature of the mixture : 21.1° C

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in(gained) = Q out(lost)

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Q ethanol=Q water

mass ethanol=

$\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g$

mass water =

$\tt mass=1~g/ml\times 45~ml=45~g$

then the heat transfer :

$\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC$

5 0

2 years ago

A welding torch requires 602.1 L of ethylene gas at 2.77 atm. What will be the pressure of the gas if ethylene is supplied by a

Katarina [22]

PV=PV
(602.1 L)(2.77atm) = (110.6 L) (X atm)
1667.817=110.6X
15.07971971 atm = X
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8 0

1 year ago

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

SVETLANKA909090 [29]

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

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2 years ago

What is the defonishon of a crystalline structure?

pogonyaev

The arrangement in space and the interatomic distances and angles of the atoms in crystals, usually determined by x-ray diffraction measurements

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2 years ago

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tino4ka555 [31]

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3 years ago

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