Question
Asked Nov 6, 2019

(100g)(4.184J/g K)(t-22)=(275g)(0.444J/g K)(t-100)

What is the step by step solution to this problem?

check_circleExpert Solution
Step 1

The given equation is

(100g)(4.184J/g K)(t-22)=(275g) (0.444J/g K)(t-100)
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(100g)(4.184J/g K)(t-22)=(275g) (0.444J/g K)(t-100)

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Step 2

The only unknown in the equation is‘t’.

This the equations of heat transfer betwee...

J
-x(t-22 K)=275gx0.444
gK
J
(100 K -t)
gK
100gx4.184-
(t-22 K) 275x0.444
(100 K-t) 100x4.184
(t-22 K) 0.2918(100 K -t)
t -22 K 29.18K -0.2918 t
(t+0.2918t)51.18 K
51.18 K
1.2918
-39.619K
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J -x(t-22 K)=275gx0.444 gK J (100 K -t) gK 100gx4.184- (t-22 K) 275x0.444 (100 K-t) 100x4.184 (t-22 K) 0.2918(100 K -t) t -22 K 29.18K -0.2918 t (t+0.2918t)51.18 K 51.18 K 1.2918 -39.619K

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