Question

10g of ice at 0 degrees absorbs 5460 J of heat energy to melt and change to water at 50 degrees C.
Calculate the specific latent heat of fusion of ice.Specific heat capacity of water is 4200 J kg^-1 K^-1

Answer 1

@Michele_Laino

Answer 2

the requested amount of heat \(Q\) to go from ice at \(0\) degrees to water at \(50\) degrees, is:
\[\huge Q = Lm + cm\Delta \theta \]
where \(m=0.01\)Kg, \(L\) is the latent heat, \(\Delta \theta=50\), \(c=4200\)

Answer 3

of course \(Q=5460\)

Answer 4

remark: (\( \Delta \theta= 50 °C= 50 °K\))

Answer 5

5460 = 0.04 times L times 2100??

Answer 6

5460 - 0.04 times L plus 2100*

Answer 7

no, please we have:
\(5460= L m+2100\)

Answer 8

so:
\[\huge L = \frac{{5460 - 2100}}{{0.010}} = ...?\]

Answer 9

336000

Answer 10

correct!

Answer 11

whats the unit..??

Answer 12

please don't forget the right unit of measure
It is \(Joules/Kg\)