A 140.0-g sample of water at 25.0°C is mixed with 108.6 g of a certain metal at 100.0°C. After thermal equilibrium is established, the (final) temperature of the mixture is 29.6°C. What is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned?

The value of specific heat of metal = 0.3537

Explanation:

Mass of water = 140 gm = 0.14 kg

Initial temperature of water = 25°c = 298 K

Mass of metal = 108.6 gm = 0.1086 kg

Initial temperature of metal = 100°c = 373 K

After thermal equilibrium the final temperature   = 29.6°c = 302.6 K

Apply energy principal

Heat lost by metal = heat gain by water  --------- (1)

⇒ Heat lost by metal = × × ( - )

⇒ Heat lost by metal = 0.1086 × × ( 373 -302.6 )

⇒ Heat lost by metal = × 7.64544  KJ -------- (2)

Now

Heat gain by water = × × ( - )

⇒ Heat gain by water = 0.14 × 4.2 × ( 302.6 - 298)

⇒ Heat gain by water = 2.7048 KJ --------------------- (3)

From the energy principal

Equation 2 = equation 3

× 7.64544  = 2.7048

= 0.3537

This is the value of specific heat of metal.

first one is the correct answer

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer