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
