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AlexFokin [52]
3 years ago
12

When 167.2 joules of heat is added to 4.00 grams of water at 10c the resulting temperature is what

Chemistry
1 answer:
Katena32 [7]3 years ago
5 0
Q=mc \Delta T \\\\
Q=168,2J\\
m=4g\\
c=4.184\frac{J}{g^{o}C}\\\\
\Delta T=\frac{Q}{mc}=\frac{167,2J}{4g*4,184\frac{J}{g^{o}C}}=9,99^{o}C\\\\
\Delta T = T_{f}-T_{i}\\\\
9,99^{o}C=T_{f}-10^{o}C\\\\
T_{f}=9,99^{o}C+10^{o}C=19,99^{o}C\approx20^{o}C
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