Question

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

Answer 1

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 2

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