Specific Heat Formula
The specific heat capacity is the quantity of heat required to change the temperature of a unit of mass (say, a gramme or a kilogramme) by one degree Celsius. Joules/kilogram/Kelvin (J/kg/K) are the standard metric units.
The heat required to increase the temperature of a particular substance’s unit mass by a given quantity is known as specific heat capacity. Calories or joules per gramme per Celsius degree are the units of specific heat.
The relationship for specific heat capacity between all quantities is expressed by the following equation given as.
Decide whether you want to warm up (add thermal energy to the sample) or cool it down (take some thermal energy away).
If you want to cool down the sample, enter a negative number for the removed energy. Let’s imagine we wish to reduce the thermal energy of the sample by 63,000 J then Q = -63,000 J in this case.
We can calculate the temperature difference between the sample’s initial and final states using the heat capacity. The difference will be negative if the sample is cooled, and positive if it is warmed.
1. A 500 gram lead cube is heated from 25 to 75 degrees Celsius. How much power did it take to heat this lead? Lead has a specific heat of 0.129 J/g°C.
Let’s start with the factors we already know.
c = 0.129 J/g°C m = 500 g
∆T = (Tfinal – Tinitial)
(75 °C – 25 °C) = 50 °C
Fill in the blanks in the above-mentioned specific heat equation.
mc∆T = Q
Q = (500 grams) (0.129 J/g°C) (50 degrees Celsius)
3225 J = Q
The lead cube was heated from 25°C to 75°C with 3225 Joules of energy.
2. Using 2330 Joules of energy, a 25-gram metal ball gets heated to 200 °C. What is the metal’s specific heat?
Solution: Make a list of all we know.
25 grammes = m
∆T = 200 degrees Celsius
2330 J = Q
Fill in the blanks in the specific heat equation.
Q = mc∆T
2330J = (25 g)c (200 °C)
(5000 g°C)c = 2330 J
Step 1: c = 0.466 J/g°C divide both sides by 5000 g°C specific heat example math
Answer: The metal’s specific heat is 0.466 J/g°C.