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August 1, 2014

Posted by **nataly** on Tuesday, October 26, 2010 at 9:50pm.

- 6th grade -
**jai**, Tuesday, October 26, 2010 at 10:19pmsimplified energy balance:

Q,absorbed + Q,lost = 0

where Q = mc(T2-T1)

in which Q=heat in J, m=mass in kg, c=specific heat capacity (which has a definite value for various substances) in J/(kg*C), T=temperature in Kelvin

*T2=final temp (the same for both and is usually the unknown)

*T1=initial temp

since in this problem, it involves only one type of substance (which is water), the c will be canceled:

Q,absorbed + Q,lost = 0

m1*c*(T2-T1,a) + m2*c*(T2-T1,b) = 0

m1*(T2-T1,a) + m2*(T2-T1,b) = 0

..from here, recall that density is mass per unit volume or:

d=m/V

density of water is approximately equal to 1 g/mL, thus,

m1 = 100 g

m2 = 100 g

since they are the same, we cancel them from the equation:

m1*(T2-T1) + m2*(T2-T1) = 0

(T2-T1,a) + (T2-T1,b) = 0

solving for T2:

T2 - 90 + T2 - 10 = 0

2(T2) = 100

T2 = 50 degrees Celsius

*note that change in T is involved thus Celsius and Kelvin unit can be used interchangeably

so there,, sorry for long explanation~

- 6th grade -
**charles**, Wednesday, October 27, 2010 at 11:58amt2=50 degrees celisius

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