View source for Specific Latent Heat of Fusion

==Key Stage 4==
===Meaning===
'''Specific Latent Heat of Fusion''' is the [[energy]] required to [[State Change|change the state]] of 1kg of a [[substance]] from a [[solid]] to a [[liquid]].

===About Specific Latent Heat of Fusion===
: The [[SI Unit]] of '''specific latent heat of fusion''' is the J/kg.
: Different [[material]]s have a different '''specific latent heat of fusion'''.
: '''Specific latent heat of fusion''' depends on the strength of the [[bond]]s holding the [[particle]]s in fixed positions in the [[solid]].
: The '''specific latent heat of fusion''' of a [[material]] can be found by [[measuring]] the [[energy]] needed to [[melting|melt]] 1kg of the [[material]].

{| class="wikitable"
|[[File:MeltingGraph.png|center|500px]]
|-
| style="height:20px; width:500px; 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]].
|}

===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_f = \frac{E}{m}</math>

Where:

<math>L_f</math> = The [[Specific Latent Heat of Fusion]] 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;" |650J of [[energy]] is needed to [[melting|melt]] 11g of solder at its [[Melting Point|melting point]]. Calculate the [[Specific Latent Heat of Fusion|'''specific latent heat''' of fusion]] of the solder correct to two [[Significant Figures|significant figures]].
| 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:left;" |'''1. State the known quantities in [[SI Unit]]s'''

E = 650J

m = 11g = 11x10<sup>-3</sup>kg
| 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;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''

<math>L_f = \frac{E}{m}</math>

<math>L_f = \frac{650}{11 \times 10^{-3}}</math>

<math>L_f = 59090 J/kg</math>

<math>L_f \approx 59000 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_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 3.4 \times 10^5 J/kg</math>
|}