Answer:
the mass of the water is 29.4 g
Explanation:
given information:
mass of iron, [tex]m_{i}[/tex] = 32.5 g
iron's temperature, [tex]T_{i}[/tex] = 21.8°C
water's temperature, [tex]T_{w}[/tex] = 63.3°C
thermal equilibrium, [tex]T_{eq}[/tex] = 58.9°C
since water and iron is being mixed, q of the water and the iron are equal, thus
[tex]q_{i}[/tex] = - [tex]q_{w}[/tex], heat being transfer
[tex]m_{i}[/tex] [tex]C_{i}[/tex] Δ[tex]T_{i}[/tex] =tex]m_{w}[/tex] [tex]C_{w}[/tex] Δ[tex]T_{w}[/tex]
(32.5)(0.449)(58.9-21.8) = tex]m_{w}[/tex] (4.18)(58.9-63.3)
tex]m_{w}[/tex] = (32.5)(0.449)(58.9-21.8)/(4.18)(58.9-63.3)
= 29.4 g
Answer: Mass of water is 29.50g
Explanation:
Since the system is well insulated the heat lost to the surrounding is assumed to be zero(0)
Therefore,
Heat gained by iron = Heat lost by the water
Mi×Ci×∆Ti = Mw×Cw×∆Tw .... eqn1
Where Mi and Mw = mass of iron and water respectively
Ci and Cw = specific heat capacity of iron and water respectively
∆Ti and ∆Tw = change in temperature of iron and water respectively
Mi= 32.5g, Mw= ? , Ci = 0.45J/gC, Cw= 4.18J/gC,
∆Ti = 58.9-21.8 = 37.1°C
∆Tw = 63.3-58.9 = 4.4°C
From eqn1
Mw = (Mi×Ci×∆Ti)/(Cw×∆Tw)
Mw = ( 32.5 × 0.45 × 37.1)/ ( 4.18 × 4.4)
Mw = 29.50g
