Respuesta :
Answer:
Te = 23.4 °C
Explanation:
Given:-
- The mass of iron horseshoe, m = 1.50 kg
- The initial temperature of horseshoe, Ti_h = 550°C
- The specific heat capacity of iron, ci = 448 J/kgC
- The mass of water, M = 25 kg
- The initial temperature of water, Ti_w = 20°C
- The specific heat capacity of water, cw = 4186 J/kgC
Find:-
What is the final temperature of the water–horseshoe system?
Solution:-
- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:
m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )
- Solve for (Te):
m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )
Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]
- Plug in the values and evaluate (Te):
Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]
Te = 2462600 / 105322
Te = 23.4 °C
Answer:
Final temperature = 23.4 °C
Explanation:
The idea here is that the heat lost by the metal will be equal to the heat gained by the water.
In order to be able to calculate the final temperature of the iron + water system, we need to know the specific heats of water and iron which are;
C_water = 4.18 J/g°C
C_iron = 0.45 J/g°C
The formula to determine the determine the amount of heat lost or gained is given by;
q = m•c•ΔT
q = heat lost or gained
m = the mass of the sample
c = specific heat of the substance
ΔT = the change in temperature, defined as final temperature minus initial temperature
Now, -q_water = q_iron
The negative sign is used here because heat lost carries a negative sign. Let's say that the final temperature of the iron + water system will be T_f
Thus, we can say that the changes in temperature for the iron and for the water will be;
ΔT_iron = T_f − 550 °C
and ΔT_water = T_f − 20 °C
This means that we will now have;
−m_iron•c_iron•ΔT_iron = m_water•c_water•ΔT_water
This now gives us ;
−m_iron•c_iron•(T_f − 550°C) = m_water•c_water•(T_f − 20°C) - - - - (eq1)
Notice that the specific heats for these two substances is given per gram. This means that you will have to
Mass of iron = 1.5 kg = 1500g
Mass of water = 25 kg = 25000 g
Plugging the relevant values into eq(1),we have;
−(1500g)•(0.45 J/g°C)•(T_f − 550°C) = (25000 g)•(4.18 J/g°C)•(T_f − 20°C)
Multiplying out to get;
-675(T_f − 550) = 104,500(T_f − 20)
-675T_f + 371,250 = 104,500T_f - 2090000
371,250 + 2090000 = 104,500T_f + 675T_f
2461250 = 105175T_f
T_f = 2461250/105175 = 23.4 °C
