Question

What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?

Answer 1

The correct approach is "12.25°C".

Step-by-step explanation:

Given:

Mass of lead,

mc = 245 g

Initial temperature,

tc = 300°C

Mass of Aluminum,

ma = 150 g

Initial temperature,

ta = 12.0°C

Mass of water,

mw = 820 g

Initial temperature,

tw = 12.0°C

Now,

The heat received in equivalent to heat given by copper.

The quantity of heat =
`m* s* t \ J`

then,

⇒
`245* .013* (300-T) = 150* .9* (T-12.0) + 820* 4.2* (T-12.0)`

⇒
`3.185(300-T) = 135(T-12.0) + 3444(T-12.0)`

⇒
`955.5-3.185T=135T-1620+3444T-41328`

⇒
`43903.5 = 3582.185 T`

⇒
`T = 12.25^(\circ) C`