How much heat ΔQ is consumed to raise the ice temperature

AI Thread Summary
To raise the temperature of ice from -20°C to 0°C, the heat required (ΔQ) is calculated using the specific heat capacity of ice, which is approximately 2.1 kJ/kg·K. The discussion reveals a misunderstanding regarding the mass of ice and the heat capacity used, with initial calculations yielding an incorrect energy value of 2.49 * 10^-21 J. The correct mass of ice is estimated at 0.0715 kg, leading to a revised heat requirement of about 6 J. The time to achieve this heat transfer at a power output of 300 W is calculated to be around 2 minutes, but the methodology and units used in the calculations were questioned for accuracy.
Firben
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1. How much heat ΔQ is consumed to raise the ice temperature from Ti = -20 Celsius to meltingpoint Tf = 0 ? how long does it take ?



2. dQ/dt = H = 300 W

dQ = cmΔT



3 I got it to be 2.49 * 10^-21 J. Is it right ?
 
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Firben said:
1. How much heat ΔQ is consumed to raise the ice temperature from Ti = -20 Celsius to meltingpoint Tf = 0 ? how long does it take ?



2. dQ/dt = H = 300 W

dQ = cmΔT



3 I got it to be 2.49 * 10^-21 J. Is it right ?
You will have to provide a bit more information. How much ice are we talking about? Does the heat source provide 300 W?

AM
 


I don't know anything about the mass.

c = 1/m * dQ/dt (1)

c = 4190 kJ/kg (heat capacity of water)

Q = mcΔT (2)

From equation (1): m = (dQ/dt)/c (3)

m = 300W/4190 k//kg = 0.0715 kg

(2) Q = (0.0715 kg)(4190 kJ/kg)(273-15-253.15)K <==>

Q = 5.99 =~ 6 J

t = Q/H (4) (heat/power)

t(time) = 6.0 J/300 W = 0.02 J/s = 2 min

Is it right ?
 


How did you get the equation (1)?
 


I just solved it, an oversight
 


The units are wrong. c has units of energy/mass. 1/m(dQ/dt) has units of energy/(mass x time).

Your derivation of m makes no sense. You are using the wrong heat capacity. You have to use the heat capacity of ice, not liquid water. Are you sure you got it solved?

AM
 

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