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g100num [7]
3 years ago
7

A 152 g sample of ice at –37oC is heated until it turns into liquid water at 0oC. Find the change in heat content of the system.

Chemistry
2 answers:
VladimirAG [237]3 years ago
5 0

Answer:

What do you mean? The Q is a bit vague

Leona [35]3 years ago
3 0

Answer:

The change in heat content is

26,000 J

.

Explanation:

Use the following equation:

Q

=

m

c

Δ

T

,

where

Q

is heat energy in Joules,

m

is mass,

c

is specific heat capacity, and

Δ

T

is change in temperature,

(

T

final

−

T

initial

)

.

Organize your data

.

Given/Known

m

=

445 g

c

ice

=

2.03 J

g

⋅

∘

C

T

initial

=

−

58

∘

C

T

final

=

−

29

∘

C

Δ

T

=

−

29

∘

C

−

(

−

58

∘

C

)

=

29

∘

C

Explanation:

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Answer:

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Explanation:

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Data obtained from the question include:

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n2 =?

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Cross multiply to express in linear form

4.11 x n2 = 2.51 x 16.9

Divide both side by 4.11

n2 = (2.51 x 16.9) / 4.11

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