Answer: The heat required by iron is 1307.11 Joules or 312.4 Cal.
Explanation:
To calculate the amount of heat released or absorbed, we use the equation:
[tex]q=mc\Delta T[/tex]
where,
q = heat absorbed
m = mass of ice = 28.4 g
c = specific heat capacity of ice = 0.5Cal/g.°C
[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=-1^oC-(-23^oC)=22^oC[/tex]
Putting values in above equation, we get:
[tex]q=28.4g\times 0.5Cal/g.^oC\times (22^oC)\\\\q=312.4Cal[/tex]
Converting the value from calories to joules, we use the conversion factor:
1 J = 0.239 Cal
So, [tex]312.4Cal=\frac{1J}{0.239Cal}\times 312.4Cal=1307.11J[/tex]
Hence, the heat required by ice is 1307.11 Joules or 312.4 Cal.
