In: Physics
A 0.420 kg iron horseshoe that is initially at 450°C is
dropped into a bucket containing 19.7 kg of water at
22.6°C. What is the final equilibrium temperature (in °C)? Neglect
any heat transfer to or from the surroundings. Do not enter
units.
Since there is no heat transferred from surroundings heat gained by system is zero
Here heat will be transferred from iron to water
Let heat energy gained by water be Q1
Let heat lost by iron be Q2
Let the temperature of water be t1 = 22.60 C
Let the temperature of iron be t2 = 4500 C
Since no heat is transferred from surroundings, heat lost by iron is equal to heat gained by water
So, Q1 = -Q2 (-ve sign represent the loss of heat)
Q1 + Q2 = 0
then m1* s1 * t1 +
m2 * s2 *
t2 =
0
where m1 is the mass of water = 19.7 kg
s1 is the specific heat of water = 4190 J/kg K
m2 is mass of iron = 0.42 kg
s2 is the specific heat of iron = 470 J/kg K
t1 =
Tf - 22.6 ,
t2 =
Tf - 450 where Tf is the final
temperature
So, 19.7 * 4190 * (Tf - 22.6) + 0.42 * 470 * (Tf - 450) = 0
82543 * Tf - 1865471.8 + 197.4 * Tf - 88830 = 0
82740 * Tf - 1954301.8 = 0
So, Tf = 1954301.8/82740.4 = 23.610 C
so final temperature is Tf = 23.610C
Here since mass of water is too high compared to iron,so final temperature is near to temperature of water