Variables
What does Q stand for?
Heat
This is the formula that includes heat, specific heat, mass, and temperature change
Q=mcΔT
A 100 g sample of a sample is heated from 20°C to 25°C. The specific heat of the sample is 4.0 J/g·°C. Calculate the heat absorbed, q.
2000 J
What is the change in temperature of an object whose initial temperature is 40 °C and reaches a final temperature of 65 °C?
15 °C
A 6.5 g object absorbs 900J of heat while its temperature increases from 300C to 770C. Calculate the specific heat of the object.
2.95 J/g0C
How do you calculate ΔT?
Final temperature - Initial Temperature
A sample absorbs 4,000 J of heat. Its temperature increases by 5°C. The substance has a specific heat of 4.0 J/g·°C. Calculate the mass of the sample.
200 g
A 300 g metal sample warms from 10°C to 20°C. The metal’s specific heat is 2.0 J/g·°C. Calculate the heat absorbed, q.
6,000 J
A 100 g sample of water absorbs 2,000 J of heat. The specific heat of water is 4.0 J/g·°C. Calculate the change in temperature (ΔT).
Answer: 5°C
q=mcΔT
2000=100*4*ΔT
2000=400*ΔT
2000/400=400/400 * ΔT
5=ΔT
(ΔT = q ÷ mc = 2000 ÷ (100 × 4.0))
A 100 g piece of lead absorbed 8900 J of heat, and the temperature started at 420C. What is the final temperature? (Hint: Lead's specific heat is 0.129 J/g0C)
731.90C is the final temperature
What is the unit for mass?
grams
A metal absorbs 4500 J of heat and warms by 15°C. The metal’s specific heat is 3.0 J/g·°C. Calculate the mass of the metal.
100 g
How much heat must be absorbed by 400 grams of water to raise its temperature by 25°C? (The specific heat of water is 4.18 J/g°C.)
41800 J
A 200 g metal sample at 20°C absorbs 4,000 J of heat. The metal’s specific heat is 2.0 J/g·°C. Calculate the final temperature of the metal.
30°C
q=mc(Tfinal-Tinitial)
4000=200*2*(Tfinal-20)
4000=400*(Tfinal-20)
4000/400=400/400*(Tfinal-20)
10=Tfinal-20
30=Tfinal
A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of the metal?
0.466 J/g°C
What does c stand for?
A 300 g sample absorbs 6,000 J of heat. Its temperature changes by 10°C. Calculate the specific heat of the substance.
2 J/g·°C
A 500 g substance absorbs heat as its temperature increases from 18°C to 30°C. Its specific heat is 2.0 J/g·°C. Calculate the heat absorbed, q.
12000 J
A 250 g substance absorbs 5,000 J of heat and ends at 30°C. Its specific heat is 2.0 J/g·°C. Calculate the initial temperature of the substance.
20°C
q=mc(Tfinal-Tinitial)
5000=250*2*(30-Tfinal)
5000=500*(30-Tfinal)
5000/500=500/500*(30-Tfinal)
10=30-Tfinal
10-30=-Tfinal
-20=-Tfinal
20 = Tfinal
How many grams of water would require 2200 J of heat to raise its temperature from 34°C to 100°C? The specific heat of water is 4.18 J/g∙C
7.97 g
What is the unit for specific heat?
J/g℃
A 400 g solid absorbs 32,000 J of heat while its temperature increases from 20°C to 40°C. Calculate the specific heat of the solid.
4 J/g·°C
A 400 g solid is heated from 15°C to 35°C. The solid has a specific heat of 2.5 J/g·°C. Calculate the heat absorbed, q.
20,000 J
A 400 g solid releases 8,000 J of heat as it cools. The solid’s specific heat is 2.0 J/g·°C. Its final temperature is 25°C. Calculate the initial temperature of the solid.
35°C
20°C
q=mc(Tfinal-Tinitial)
8000=400*2*(25-Tfinal)
8000=800*(25-Tfinal)
8000/800=800/800*(25-Tfinal)
10=25-Tfinal
10-25=-Tfinal
15= Tfinal
A 55 kg block of metal has an original temperature of 15.0°C and 0.45 J/g∙°C. What will be the final temperature of this metal if 450 J of heat energy are added?
33.20C is the final temperature