How much heat, in oules and in, calories, is required to heat a 28.4-g ( 1-oz) ice cube from -23.0 to -1.0 degree celcius?

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:

where,

q = heat absorbed

m = mass of ice = 28.4 g

c = specific heat capacity of ice = 0.5Cal/g.°C

= change in temperature =

Putting values in above equation, we get:

Converting the value from calories to joules, we use the conversion factor:

1 J = 0.239 Cal

So,

Hence, the heat required by ice is 1307.11 Joules or 312.4 Cal.