Question

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

Answer 1

$Q=mc \Delta T \\\\ Q=168,2J\\ m=4g\\ c=4.184\frac{J}{g^{o}C}\\\\ \Delta T=\frac{Q}{mc}=\frac{167,2J}{4g*4,184\frac{J}{g^{o}C}}=9,99^{o}C\\\\ \Delta T = T_{f}-T_{i}\\\\ 9,99^{o}C=T_{f}-10^{o}C\\\\ T_{f}=9,99^{o}C+10^{o}C=19,99^{o}C\approx20^{o}C$