SNC 2DHeat Problems – (in a state only)
Heat problems deal with a mathematical analysis of the amount of heat required to raise the temperature of a substance. A substance’s unique heating property is reflected in the specific heat capacity whose symbol is “c”. ( c is different for each substance and is often subscripted with the material – cwater ) “c” values can be found in tables such as the one on page 427 of your textbook.
The other factors in heat problems are mass (m), temperature change (∆T), and heat (Q). The standard units for these variables are kg, oC, and J for Joules. These units can vary by adding prefixes to the unit – i.e.kJ. Students must check the units accordingly either through unit analysis or at the beginning of the problem to ensure the units all agree.
The equation for Heat problems during a state (solid, liquid, gas) is
Q = m∙c∙∆T
∆T can often be expanded to T2-T1 when the question gives two temperatures.
Sample problem
Determine the quantity of heat required to raise the temperature of 20 g of water by 40 oC.
Q = ∆T = T2-T1Q = m∙c∙∆T
m =20 g
cwater =4.18 J/goC
∆T = 40 oC
Or
T2 =
T1 =
Problems
- Determine the amount of heat required to raise the temperature of 20 g of brick by 40 oC.
Q = ∆T = T2-T1 Q = m∙c∙∆T
m = 20 g= (20 g)( 0.84 J/goC)(40 oC)
c = 0.84 J/goC= 672 J
∆T = 40 oC
Or
T2 =
T1 =
- Determine the amount of heat required to raise the temperature of 1.50 kg of brick by 40 oC.
Q = ∆T = T2-T1Q = m∙c∙∆T
m = 1.50 kg = 1 500 g = (1500 g) (0.84 J/goC)(40 oC)
c = 0.84 J/goC = 5040 J
∆T = 40 oC
Or
T2 =
T1 =
- Determine the amount of heat required to raise the temperature of 50 g of wet mud from 20 oC to 80 oC.
Q = ∆T = T2-T1Q = m∙c∙∆T
m = 50 g = 80oC - 20oC = (50g)(2.51 J/goC)(60 oC)
c = 2.51 J/goC = 60 oC = 7530 J
∆T =
Or
T2 = 80 oC
T1 = 20 oC
- Determine the amount of heat required to lower the temperature of 50 g of wet mud from 80 oC to 20 oC.
Q = ∆T = T2-T1 Q = m∙c∙∆T
m = 50 g = 20oC - 80oC = (50g)(2.51 J/goC)(-60 oC)
c = 2.51 J/goC = -60 oC = -7530 J
∆T =
Or
T2 =
T1 =
- If 7 500 J of heat is applied to 60 g of water, by how much would the temperature change?
Q = 7 500 J∆T = T2-T1Q = m∙c∙∆T
m = 60 g7 500 J = (60g)(4.18 J/goC)∆T
c = 4.18 J/goC7 500 J = 250 J/oC)∆T
∆T = 7 500 J/ 250 J/oC)=∆T
Or ∆T = 30 oC
T2 =
T1 =
- If 9.5 kJ of heat is applied to 160 g of water, and the water is at 20 oC, what would the final temperature be?
Q = 9.5 kJ = 9 500 J ∆T = T2-T1Q = m∙c∙∆T
m = 160 g14.2 oC = T2-20oC9 500 J = (160g)(4.18 J/goC)∆T
c = 4.18 J/goC14.2 oC + 20oC = T29 500 J = 668.8 J/oC)∆T
∆T = 34.2 oC = T29 500 J/ 668.8 J/oC)=∆T
Or ∆T = 14.2 oC
T2 =
T1 = 20 oC
- If 9.5 kJ of heat is removed from 160 g of granite, and the granite is at 120 oC, what would the final temperature be?
Q = -9.5 kJ = -9 500 J ∆T = T2-T1Q = m∙c∙∆T
m = 160 g-75.2 oC = T2–120oC-9 500 J = (160g)(0.79 J/goC)∆T
c = 0.79 J/goC-75.2 oC+120oC = T2-9 500 J = 126.4 J/oC)∆T
∆T = 44.8oC = T2-9 500 J/ 126.4 J/oC)=∆T
Or ∆T = -75.2 oC
T2 =
T1=120 oCNote – because heat is removed, Q is negative.