chemistry chemistry

Question

The specific heat of a certain type of cooking oil is 1.75 cal/(g ·°C). How much heat energy is needed to raise the temperature of 2.56 kg  of this oil from  23 C to 191 C? 

Answer 1

Thermodynamics

The relationship you need is q = mcΔT where q is heat (in cal or Joules), m is mass (in g), c is specific heat, and ΔT is the change in temperature (in degrees C or K)

You are given: c = 1.75 ^cal/_(g•°C)

m = 2.56 kg; convert to g, 2.56 kg • ^{1000 g} / _{1 kg} = 2560 g

T_initial = 23°C

T_final = 191°C

ΔT = T_final – T_initial = 191°C – 23°C = 168°C

q = mcΔT

q = (2560g)(1.75 ^cal/_{(g•°C))(168°C)}

q = 752 640 cal