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HEAT: Manipulating latent heat equations

Total questions: 11

Worksheet time: 6mins

Name
Class
Date
1.

This formula is used to find the amount of heat (energy) neccesary when there's a phase change

a)

Q= (h)(ΔT)

b)

Q=(m)(L)

c)

Q=(m)(c)(ΔT)

d)

Q=(k)(A)(dT)

2.

What is the formula to calculate specific latent heat of vaporisation?

a)

L = Q/m

b)

c = Q/(mΔƟ)

c)

Q = mcΔƟ

d)

Q = mL

3.

the symbol for specific latent heat is

a)

c

b)

L

c)

Q

d)

m

4.
What is the equation to measure change in Thermal Energy?
a)
Q=mc∆t
b)
Q=mc
c)
Q= ∆mct
d)
m=QC
5.
What would be the effect on the particles if more heat is supplied to the system?
a)
They would slow down
b)
They would stop moving
c)
They would speed up
d)
There would be no effect
6.

How much heat is needed to change 12 grams of ice at 0 ºC to 12 grams of water at 0ºC? Heat of fusion = 334 kJ/kg; Specific heat of water = 4186 J/kgoC

a)

Q = 12 x 4186 x 0

b)

Q = 12 x 334

c)

Q = 12000 x 334

d)

Q = 0.012 x 334

7.

How much heat must be lost to change of 110 grams of water at 0 ºC to 110 grams of ice at 0ºC? Heat of fusion = 334 kJ/kg; Specific heat of water = 4186 J/kgoC

a)

Q = 110 x 4186

b)

Q = 110000 x 4186

c)

Q = 0.110 x 334

d)

Q = 110 x 334

8.
What  is the formula to calculate heat energy required to raise the temperature of any substance?
a)
Q = m Cp ∆T
b)
Q = m Cp
c)
Q =  ½ m v
d)
m = Q Cp
9.

If the specific heat of water is 4,186 J/kg°C, how much heat is required to increase the temperature of 1.2 kg of water from 23 °C to 39 °C?

a)

Q = 1.2 x 4186

b)

Q = 1.2 x 4186 x 7

c)

Q = 1.2 x 4186 x 17

d)

Q = 0.0012 x 4186 x 17

10.
Using the heat equation, what would the formula look like if we were solving for change in temperature?
a)
Q m = ∆T Cp
b)
Q / (m Cp)  =  ∆T
c)
Q m / Cp  =  ∆T
d)
m c Q  =  ∆T
11.
What unit do you use to measure Thermal Energy?
a)
J/Kg ºC
b)
Kg
c)
ºC
d)
J