Verified answer

Mass of silver = M = 17.50 kg

Initial temperature of silver = t1 = 21º C

Melting point of silver = tm = 961.93º C

Specific heat capacity of silver = = Cp = 0.0556 ca/g-deg = (55.6)*(4.19) J/kg-deg

Latent heat of fusion of silver = Lf = 25 cal/g = 25*4.19*10^3 J/kg

Quantity of Heat required = MCp(tm - t1) + MLf = M{Cp(tm - t1) + Lf}

= (17.50){(55.6)(4.19)(961.93 - 21) + 25000*4.19} J

= (17.50){(55.6)*(940.93) + 25000}*(4.19) J - Now calculate and find out.

Melting point of silver = 962° C

https://www.google.com/search?source=ig&hl=en&rlz=...

Specific heat of silver = 0.0558 cal/(g-K)

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/...

Latent heat of fusion of silver = 26.5 cal/g

http://www.engineeringtoolbox.com/fusion-heat-meta...

Using the above data, heat required to melt 17.50 kg of silver at 21° C at atm. pressure

= 17500 * [0.0558 (962 - 21) + 26.5] cal

= 1382636.5 cal

= 1382.6365 Kcal.

warmth extra = specific warmth x mass x ( T very final - T preliminary) 22 C = 295.15 ok Melting element of silver = 1234.ninety 3 ok warmth extra = (.233 J/gram*ok) x 15,3 hundred grams x ( 1234.ninety 3 ok- 295.15K) Grams and ok cancel one yet another out your left with Joules warmth extra = 3,350,222 J

the stronger the force the more heat will be needed