Question

. A 0.0255 kg silver ring (Cp= 234 J kg-1 ºC-1 ) is heated to 84 ºC and then placed in a calorimeter containing 0.05 kg of water at 24 ºC. The calorimeter is not perfectly insulated, so 0.14 kJ of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?

Answer 1

T = 37.27°C

Step-by-step explanation:

We are given;

Mass of silver ring; m_s = 0.0255 kg

Specific heat capacity;Cp= 234 J kg-1 ºC-1

Silver is heated up to a temperature of;T_s1 = 84°

Mass of water;m_w = 0.05 kg

Temperature of water;T_w1 = 24°

Heat transferred to surrounding = 0.14 KJ

Specific heat capacity of water = 4200 J/kg°C

Applying the law of conservation of energy, we have;

Heat from silver cup = Heat gained by calorimeter + heat transferred to the surroundings

Let final temperature be "T"

Formula for heat is Q = m•cp•Δt

Thus;

(0.0255)(234)(84 - T) = [(0.05)(4200)(T - 24)] + 0.14

Expanding, we have;

5.967(84 - T) = 21(T - 24) + 0.14

501.228 - 5.967T = 21T - 504 + 0.14

501.228 + 504 - 0.14 = 21T + 5.967T

1005.088 = 26.967T

T = 1005.088/26.967

T = 37.27°C