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.
