Question

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

Answer 1

The correct answer is Option c (0.163 g).

Step-by-step 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,

⇒
`Q=mc \Delta T`

By substituting the values, we get

`0.21=m* 4.184* 0.308`

`m=(0.21)/(0.308* 4.184)`

`=(0.21)/(1.28867)`

`=0.163 \ g`