Question

Suppose a 14-gram sample of iron is heated from 20.0°C to 25.0°C. The specific heat of iron is 0.11 cal/g°C. How much heat energy was absorbed by the iron? 38.5 cal 7.7 cal 636 cal 69.3 cal

Answer 1

I believe The Answer To This question Is


7.7 calories


hope i was helpful

Answer 2

Answer : The heat energy absorbed by the iron is 7.7 cal.

Explanation :

Formula used :

`Q=m* c* \Delta T\\\\Q=m* c* (T_2-T_1)`

where,

Q = heat absorb = ?

m = mass of iron = 14 g

c = specific heat of iron =
`0.11Cal/g^oC`

`\Delta T` = change in temperature

`T_1` = initial temperature =
`20.0^oC`

`T_2` = final temperature =
`25.0^oC`

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

`Q=14g* 0.11Cal/g^oC* (25.0-20.0)^oC`

`Q=7.7Cal`

Therefore, the heat energy absorbed by the iron is 7.7 cal.