If 1000 calories of heat is added to a 50.00g sample of water at 20.0 degrees C, what would be the final temp?

Question

of the water sample?

Answer 1

1000 cal/(50.00g x 1 cal/g'C) = dT
20.0 C' = dT
dT = T2 - T1
T2 - 20.0'C = 20.0C'
T2 = 40'C

Answer 2

the rise of temperature is

Q= mcdT
where m mass in gram, c specific heat for water 1 calorie/degree Celsius and dT rise in temperature so
dT= Q/mc= 1000/50*1 =20°C

and the final temperature is 20+20 .0 degreesC