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.

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be