Answer to Question #311247 in Mechanical Engineering for syrus

Question

What is the temperature of 2 liters of water at 30°C after 3,500 calories of heat have been added?Use: Cp=4.187 KJ/Kg.-°K, 1 Kg. = 1 liter, 1 Kcal = 4.18 KJ

Answer 1

$Q=mc_p \Delta T$

$=mc_p(T_2-T_1)$

$0.500 Kcal (4.187 KJ/Kcal) = 2 liters (1kg/liter)(4.187 KJ/kg-°C)(t2–30°C)$

$T_2=30.25\degree C$