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If the temp. of 63.4 grams of water changes from 22.3C to 28.7C, how many joules of heat are involved? Show work

Sagot :

Answer:

Q = 1698.51 J

Explanation:

Given that,

Mass of water, m = 63.4 grams

It changes temperature from 22.3C to 28.7C.

We need to find the heat involved. The heat involved due to the change in temperature is given by :

[tex]Q=mc\Delta T[/tex]

Where

c is the specific heat of water, c = 4.186 joule/gram °C

So,

[tex]Q=63.4 \times 4.186 \times (28.7- 22.3)\\\\Q=1698.51\ J[/tex]

So, 1698.51 J of heat is involved.

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