general chemistry

Question

The specific heat of a certain type of cooking oil is 0.418 cal/(g·°C). How much heat energy is needed to raise the temperature of 0.959 kg of this oil from 23.0 °C to 60.0 °C?

Answer 1

For this question you can use the formula q = mCΔT where

q = heat = ?

m = mass = 0.959 kg = 959 g

C = specific heat = 0.418 cal/g/deg

ΔT = change in temperature = 60.0 - 23.0 = 37.0 degrees

q = (959 g)(0.418 cal/g/deg)(37.0 deg) = 14832 cal = 14,800 cal

This is also equivalent to 14.8kcal (3sig. figs.)