# Heat Transfer and Specific Heat Fill in Note Sheet With Answered Examples

```Name: ___braian tello______________________
_________________
Period: _7th___
Date:
Heat Transfer
When objects of different temperature are brought
together, and heat is transferred from the highertemperature objects to the lower-temperature
objects, the total internal energy is conserved.
Applying conservation of energy means that the
total heat transferred from the hotter objects must
equal the total heat transferred to the cooler
objects. If the temperature of an object changes,
the heat (Q) added or removed can be found using
the equation:
SPECIFIC HEAT (c) is a measure of ____1.00c__________________________________
Specific heat is defined as ____ the heat required to raise the temperature of the unit mass of a
given substance by a given amount (usually one degree).
If specific heat is high, the substance will stay ____hot________________________
Write the equations here:
Identify the variables and their units here:
q = mc∆T. Q
m = heat transfer
c = Celsius
Tf =
ΔT = Thermal
Ti =
Ex 1 How much heat energy is required to raise the temperature of 44.0 grams of lead from
room temperature, 21oC, to its melting point, 327oC? The specific heat of lead is 0.13
J/g*oC.
m=44g, Ti= 21oC, Tf= 327oC, c= 0.13J/goC, q=?
∆ T= Tf-Ti
= 327 – 21 = 306oC
q = mc∆T
= 44x 0.13x 306= 1750.32J
Ex 2 In a laboratory experiment, 58.1 grams of copper lose 1.6x103 joules in cooling from
99.8oC to 28.4oC. Calculate the specific heat of copper.
m= 58.1g q= 1.6x103J, Ti = 99.8oC, Tf= 28.4oC. c=?
∆T = Tf – Ti = 28.4 – 99.8 = - 71.4oC (always take ∆T in absolute value).
q= m c ∆T
1.6x103 = 58.1 x c x 71.4
1600 = 4148.34 c  C= 1600/ 4148.34 = 0.386J/goC
```