Question

1050 calories of heat from soybean oil was applied to a 250 g sample of liquid water with an initial temp of 25 degrees C. Determine: a) the final temp of the water, and, b) the change in temperature.

Answer 1

heat = 1050 cal

mass = m = 250 g

Cp of water = 4.184 J / g oC

Q = m Cp dT

1050 = 250 x 1 x dT

dT = 4.2 oC

b)  the change in temperature = 4.2 ^oC

a) dT = T2 - T1 = 4.2

T2 -25 = 4.2

T2 = 29.2 oC

final temperature = 29.2 ^oC