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.
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.
Per hour-68x60=4080
Per day-4080x24=97920
Per week-97920x7=685440
Energy were released from the walnut, q = 1,673.6 J
<h3>Equation :</h3>
To find the energy using formula,
q = mcΔt
where,
q is charge
m is mass
c is specific heat of water
Δt is change in temperature
So, given
t₁ = 50°C
t₂ = 60°C
m = 40g
c = 4.184 J/g
Now putting the values known,
We get,
q = mc(t₂ - t₁)
q = 40g x 4.184 J/g x (60 - 50)
q = 167.36 J x 10
q = 1,673.6 J
<h3>What is heat energy?</h3>
Heat is the thermal energy that is transferred when two systems with different surface temperatures come into contact. Heat is denoted by the letters q or Q and is measured in Joules.
To know more about specific heat :
brainly.com/question/11297584
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Atoms<span> are </span>arranged in molecular compounds in groups.
<span>
For </span>covalent compounds<span>: </span>
<span>consider drawing the lewis structure of the covalent compound in question, putting the atom which is least electronegative (save hydrogen) in the center.
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