How much heat ΔQ is consumed to raise the ice temperature

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Homework Help Overview

The discussion revolves around calculating the heat consumed to raise the temperature of ice from -20 Celsius to its melting point at 0 Celsius, along with inquiries about the time required for this process.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between heat transfer, mass, and temperature change using equations related to heat capacity. Questions arise regarding the mass of the ice and the appropriate heat capacity to use.

Discussion Status

Some participants have provided calculations and attempted to derive the mass of the ice and the heat required. However, there are concerns about the correctness of the heat capacity used and the units in the derivations. Multiple interpretations and approaches are being discussed, with no explicit consensus reached.

Contextual Notes

There is a lack of information regarding the mass of the ice, and participants are questioning the assumptions made about the heat capacity, specifically whether to use that of ice or liquid water.

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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