Attempt 2 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 2.50 kg of this oil from 23 ∘C to 191 ∘C?

Question

I need the answer in Cal

Answer 1

The relevant equation for this problem can be written as Q = s x m x Δt where Q is the heat energy required, s is the specific heat, m is the mass of the oil in grams, and Δt is the change in temperature.

Given: s = 1.75 cal/(g•ºC)

m = 2.50 kg x 1000 g/kg = 2500 grams

Δt = (191 - 23)ºC = 168ºC

Substitute these values in the equation above to get Q = 735000 calories or 735 kcal