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If 0.21J of heat cause a 0.308 degree C temperature change, what mass of water is present?
a 0.0702 g
b 0.00540 g
c 0.163 g
d 18.4 g

Sagot :

Cricetus

Answer:

The correct answer is Option c (0.163 g).

Explanation:

Given:

Heat energy,

Q = 0.21 J

Specific heat,

c = 4.184 J/g°c

Change in temperature,

ΔT = 0.308°C

As we know,

⇒ [tex]Q=mc \Delta T[/tex]

By substituting the values, we get

 [tex]0.21=m\times 4.184\times 0.308[/tex]

     [tex]m=\frac{0.21}{0.308\times 4.184}[/tex]

         [tex]=\frac{0.21}{1.28867}[/tex]

         [tex]=0.163 \ g[/tex]

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