Difference between revisions of "Specific Latent Heat"

Revision as of 13:36, 6 March 2019

Key Stage 4

Meaning

Specific Latent Heat is the energy required to change the state of 1kg of a substance.

About Specific Latent Heat

The SI Unit of specific latent heat is the J/kg.

Different materials have a different specific latent heat.

Specific latent heat depends on the strength of the bonds holding the particles to each other.

The specific latent heat of a material can be found by measuring the energy needed to change the state of the material.


The increase internal energy during the time when the temperature remains constant is the energy required to melt the material. This can be used to calculate the specific latent heat of fusion.	The increase internal energy during the time when the temperature remains constant is the energy required to vaporise the material. This can be used to calculate the specific latent heat of vaporisation.

Equation

NB: You do not need to remember this equation but you need to be able to use it.

Specific Latent Heat = (Energy Transferred)/(Mass)

\(L = \frac{E}{m}\)

Where\[L\] = The Specific Latent Heat of the material.

\(E\) = The Energy transferred to the material during the state change.

\(m\) = The mass of the material.

Example Calculations

An 11kg block of ice at 0°C is heated by an immersion heater until it completely melts. The immersion heater is connected to a Joulemeter which reads 3.7MJ. Calculate the specific latent heat of fusion of the water ice correct to two significant figures.	A pan containing 500g of water at 100°C is heated with an energy of 1.13kJ until all of the water has been vaporised. Calculate the specific latent heat of vaporisation of the water steam correct to two significant figures.
1. State the known quantities in SI Units E = 3.7MJ = 3.7x10⁶J m = 11kg	1. State the known quantities in SI Units E = 1.13kJ = 1.13x10³J m = 500g = 0.5kg
2. Substitute the numbers into the equation and solve. \(L_f = \frac{E}{m}\) \(L_f = \frac{3.7 \times 10^6}{11}\) \(L_f = 3.36363 \times 10^5 J/kg\) \(L_f \approx 3.4x10^5 J/kg\)	2. Substitute the numbers into the equation and solve. \(L_v = \frac{E}{m}\) \(L_v = \frac{1.13 \times 10^3}{0.5}\) \(L_v = 2260 J/kg\) \(L_v \approx 2300 J/kg\)

@@ Line 15: / Line 15: @@
 | style="height:20px; width:300px; text-align:left;" |The increase [[Internal Energy|internal energy]] during the [[time]] when the [[temperature]] remains constant is the [[energy]] required to [[melt]] the [[material]]. This can be used to calculate the [[Specific Latent Heat of Fusion|'''specific latent heat''' of fusion]].
 | style="height:20px; width:300px; text-align:left;" |The increase [[Internal Energy|internal energy]] during the [[time]] when the [[temperature]] remains constant is the [[energy]] required to [[boil|vaporise]] the [[material]]. This can be used to calculate the [[Specific Latent Heat of Vaporisation|'''specific latent heat''' of vaporisation]].
+|}
+===Equation===
+''NB: You do not need to remember this equation but you need to be able to use it.''
+Specific Latent Heat = (Energy Transferred)/(Mass)
+<math>L = \frac{E}{m}</math>
+Where:
+<math>L</math> = The [[Specific Latent Heat]] of the [[material]].
+<math>E</math> = The [[Energy]] [[Energy Transfer|transferred]] to the [[material]] during the [[State Change|state change]].
+<math>m</math> = The [[mass]] of the [[material]].
+===Example Calculations===
+{| class="wikitable"
+| style="height:20px; width:300px; text-align:center;" |An 11kg block of [[ice]] at 0°C is [[heat]]ed by an [[Immersion Heater|immersion heater]] until it completely [[melting|melts]]. The [[Immersion Heater|immersion heater]] is connected to a [[Joulemeter]] which reads 3.7MJ. Calculate the [[Specific Latent Heat of Fusion|'''specific latent heat''' of fusion]] of the [[water]] [[ice]] correct to two [[Significant Figures|significant figures]].
+| style="height:20px; width:300px; text-align:center;" |A pan containing 500g of [[water]] at 100°C is [[heat]]ed with an [[energy]] of 1.13kJ until all of the [[water]] has been vaporised. Calculate the [[Specific Latent Heat of Vaporisation|'''specific latent heat''' of vaporisation]] of the [[water]] [[steam]] correct to two [[Significant Figures|significant figures]].
+|-
+| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s'''
+E = 3.7MJ = 3.7x10<sup>6</sup>J
+m = 11kg
+| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s'''
+E = 1.13kJ = 1.13x10<sup>3</sup>J
+m = 500g = 0.5kg
+|-
+| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
+<math>L_f = \frac{E}{m}</math>
+<math>L_f = \frac{3.7 \times 10^6}{11}</math>
+<math>L_f = 3.36363 \times 10^5 J/kg</math>
+<math>L_f \approx 3.4x10^5 J/kg</math>
+| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
+<math>L_v = \frac{E}{m}</math>
+<math>L_v = \frac{1.13 \times 10^3}{0.5}</math>
+<math>L_v = 2260 J/kg</math>
+<math>L_v \approx 2300 J/kg</math>
 |}