How is the Calorimetry Mixture Problem Calculated?

[Bashyboy]

Hello, my problem is as follows: " If 150 g of lead at 100°C were placed in a calorimeter with 50 g of water at 28.8°C and the resulting temperature of the mixture was 22°C, what are the values of q(lead), q(water), and q(cal)? (Knowing that the specific heat of water is 4.184 J/g °C and the specific heat of lead is 0.128 J/g °C)

The part that I don't understand in the calculations is this:
q(lead) = 0.128 J/g °C x 150g x (28.8°C - 100°C) = -1.37 E3 J

Why do they subtract 100 from 28.8? From my present understanding, 100 degrees is the initial temperature of the lead and 28.8 degrees is the initial temperature of water; shouldn't they subtract 100 degrees from 22 degrees?

Thank you.

 

[Borek]

No way putting hot lead in a warm water can yield a mix colder than initial temperatures of both substances.

q is mcΔT, where ΔT=T_final-T[/sub]initial[/sub], so at least the second part (-100) is correct. It yields a negative q, meaning lead lost heat - that also makes sense, and fits the convention.