Question

How many joules are absorbed when 64.3 grams of water is heated from 40.0 °C to 74.0 °C?

Answer 1

Answer : The correct option is, (B) 9150 joules

Explanation : Given,

Initial temperature = $40.0^oC$

Final temperature = $74.0^oC$

Mass of water = 64.3 g

Specific heat of water = $4.184J/g^oC$

Formula used :

$Q=m\times C_w\times \Delta T$

or,

$Q=m\times C_w\times (T_2-T_1)$

where,

Q = heat absorbed = ?

m = mass of water

$C_w$ = specific heat of water

$T_1$ = initial temperature

$T_2$ = final temperature

Now put all the given value in the above formula, we get:

$Q=64.3g\times 4.184J/g^oC\times (74-40)^oC$

$Q=9150J$

Therefore, the heat absorbed will be, 9150 J

Answer 2

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