Question

A 255 g lead ball at a temperature of 81.6°C is placed in a light calorimeter containing 153 g of water at 22.3°C. Find the equilibrium temperature of the system.

Answer 1

32.73 Degree C

Step-by-step explanation:

mass of lead, m1 = 255 g, T1 = 81.6 degree C

mass of water, m2 = 153 g, T2 = 22.3 degree C

Let the equilibrium temperature be T.

According to the principle of caloriemetery.

heat lost by the lead = heat gained by water

m1 x c1 x decrease in temperature = m2 x c2 x increase in temperature

where, c1 and c2 be the specific heat capacity of lead and water respectively.

c1 = 0.128 cal/gm C

c2 = 1 cal/gm C

So,

255 x 0.128 x (81.6 - T) = 153 x 1 x (T - 22.3)

2663.424 - 32.64 T = 153 T - 3411.9

6075.324 = 185.64 T

T = 32.73 Degree C