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 :

$Q=m\times c\times \Delta T\\\\Q=m\times c\times (T_2-T_1)$

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:

$Q=14g\times 0.11Cal/g^oC\times (25.0-20.0)^oC$

$Q=7.7Cal$

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