1426.58 J and 340.90 calories heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C.
Step-by-step explanation:
Data given:
mass = 28.4 gram
initial temperature = -23 degrees
final temperature = 1degress
change in temperature ΔT = Tfinal - Tinitial
ΔT = 1 -(-23)
ΔT = 24 degrees
specific heat capacity of ice cube c = 2.093 J/g C
Formula used:
q = mc ΔT
putting the values in the equation:
q= 28.4 x 2.093 x 24
= 1426.58 J
ENERGY IN CALORIES:
340.90 calories is the energy is required in the process.