Question

5.the temperature of a sample of water increases from 20°c to 46.6°c as it absorbs 5650 calories of heat. what is the mass of the sample? (specific heat of water is 1.0 cal/g °c)

Answer 1

`Q=cm \Delta T \ \ \ \Rightarrow \ \ \ m=(Q)/(c \Delta T)\\\\ Q=5650cal\\ c=1(cal)/(g^(o)C)\\ \Delta T=46,6^(o)C-20^(o)C=26,6^(o)C\\\\ m=(5650cal)/(1(cal)/(g^(o)C)*26,6^(o)C)\approx 212,41g`