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

Answer:

The correct approach is "12.25°C".

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 = $m\times s\times t \ J$

then,

⇒ $245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)$

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

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

⇒                         $43903.5 = 3582.185 T$

⇒                                  $T = 12.25^{\circ} C$