Finding mass in a quanity of heat problem

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SUMMARY

The discussion focuses on calculating the mass of a copper part that loses 80 kcal of heat during a quenching process, cooling from 400 degrees Celsius to 30 degrees Celsius. The relevant equation used is Q=mcΔT, where Q represents heat lost, m is mass, c is specific heat capacity, and ΔT is the change in temperature. The user attempts to rearrange the formula to find mass, using the specific heat capacity of copper at 0.093 cal/g°C and a temperature change of 370°C. The correct formula setup is confirmed as m = Q/(cΔT).

PREREQUISITES
  • Understanding of the specific heat capacity of materials, specifically copper.
  • Familiarity with the formula Q=mcΔT for heat transfer calculations.
  • Basic algebra skills for rearranging equations.
  • Knowledge of temperature scales and conversions between Celsius and Kelvin.
NEXT STEPS
  • Review the concept of specific heat capacity and its application in thermal calculations.
  • Practice problems involving heat transfer using the formula Q=mcΔT.
  • Explore the implications of heat loss in different materials during quenching processes.
  • Learn about the effects of temperature changes on material properties in thermal treatments.
USEFUL FOR

Students studying thermodynamics, engineers involved in material processing, and anyone interested in heat transfer principles in metallurgy.

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Homework Statement



In a heat treatment, a hot copper part is quenched with water, cooling it from 400 degrees Celsius to 30 degrees Celsius. What was the mass of the part if it loses 80 kcal of heat.


Homework Equations



Q=mcΔT

The Attempt at a Solution



m=c ΔT/Q m=(0.093cal/g*C°)(370 C° )/ 80Kcal


I gave this a shot but I don't feel like it's right. Any help and hints or even just setting up the formula is appreciated big time. I have a feeling this one is kind of obvious I'm just not seeing it.
 
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Damien20 said:

Homework Statement



In a heat treatment, a hot copper part is quenched with water, cooling it from 400 degrees Celsius to 30 degrees Celsius. What was the mass of the part if it loses 80 kcal of heat.


Homework Equations



Q=mcΔT

The Attempt at a Solution



m=c ΔT/Q m=(0.093cal/g*C°)(370 C° )/ 80Kcal

m = Q/cΔT
 

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