What is the calories needed to heat 11.4 g of water from 21.0°C to 43.0°C?
You will need to use the equation below to solve this problem.
m - mass in grams of the substance
c - specific heat capacity (J/g°C) (Varies according to the state of substance)
ΔT - Change in temperature (°C)
Here, it is given that the mass of water is 11.4 grams. Meanwhile, the change in temperature would be
According to my textbook, the specific heat capacity for liquid water is 4.19J/g°C
Substitute all of these values into the equation and we will obtained
If we were to convert it into kJ and take significant figures into consideration, the answer would be
As 2,500kcal is equivalent to 10,500 kJ