Question

A 100g piece of iron at 150oC is emersed in 268.5 g of water at 20oC. the temperature of both iron and water became 25oC. If the specific heat of iron is 0.449 J/g.K, find the specific heat of water?​

Answer 1

cW = 4.2J/g.k

Step-by-step explanation:

Heat gained = Heat lost

iron(mc∆€) = water(mc∆€)

mi= 100g

ci = 0.449

∆€i= 150-25 = 125

...

mw= 268.5g

cw = ?

∆€w=25-20 = 5

...

100×0.449×125= 268.5×cw×5

5612.5= 1342.5cw

cw = 5612.5/1342.5

cw= 4.18 ~ 4.2J/g.K