1. A 0.40 kg iron horseshoe that is initially at 500 degrees Celsius is dropped into a bucket containing 20 kg of water at 22 degrees Celsius. What is the final equilibrium temperature? Neglect any energy transfer to or from the surrounding.

2. An aluminum cup contains 225 g of water and a 40 g coffee stirrer, all at 27 degrees Celsius. A 400 g sample of silver at an initial temperature of 87 degrees Celsius is placed in the water. The stirrer is used to stir the mixture gently until it reaches its final equilibrium temperature at 32 degrees Celsius. Calculate the mass of the aluminum cup.

3. A 50 g ice cube at 0 degrees Celsius is heated until 45 g has become water at 100 degrees Celsius and 5 g has become steam at 100 degrees Celsius. How much energy was added to accomplish this?

4. What mass of steam that is initially at 120 degrees Celsius is needed to warm 350 g of water and its 300 g aluminum container from 20 degrees Celsius to 50 degrees Celsius?

5. Steam at 100 degrees Celsius is added to ice at 0 degree Celsius.
a. Find the amount of ice melted and the final temperature when the mass of the steam is 10 g and the mass of ice is 50 g.
b. Repeat with steam of mass 1.0 g and ice of mass 50 g.

4. What mass of steam that is initially at 120 degrees Celsius is needed to warm 350 g of water and its 300 g aluminum container from 20 degrees Celsius to 50 degrees Celsius?

Let's start with this one.

This is a heat transfer problem where we have to equate the thermal energy lost by the steam to the thermal energy gained by the water and aluminum container.

There are 3 stages as the steam loses thermal energy. In the first stage, the steam is cooled to 120-(50-20)=90 Celsius.

The thermal energy freed is then Q_{1}=m_{s}c_{s}{\Delta}T=m_{s}(2.01\times 10^{3} \;\ J/kg\cdot C^{\circ})(30 \;\ C^{\circ})=m_{s}(6.03\times 10^{4} \;\ J/kg)

In the second stage, the steam is converted to water. To find the thermal energy removed, we use the latent heat of vaporization and Q=mL_{v}.

Q_{2}=m_{2}(2.26\times 10^{6} \;\ J/kg)

In the third stage, the temperature of the water is raised to 50 C. This frees an amount of thermal energy

Q_{3}=m_{s}c_{w}{\Delta}T=m_{s}(4.19\times 10^{3} \;\ J/kg\cdot C)(50 \;\ C)=m_{s}(2.09\times 10^{5} \;\ J/kg)

Now, if we equate the thermal energy lost by the steam to the thermal energy gained by the water and aluminum, then:

m_{s}(6.03\times 10^{4} \;\ J/kg)+m_{s}(2.26\times 10^{6} \;\ J/kg)+m_{s}(2.09\times 10^{5} \;\ J/kg)=(.35 \;\ kg)(4.10\times 10^{3} \;\ J/kg \;\ C)(30 \;\ C)+(.300 \;\ kg)(900 \;\ J/kg\cdot C)(30^{\circ}C)

Finding the mass and solving for m gives us \fbox{m_{s}=20.6 \;\ grams}


WHEW... check my solution. Easy to make a mistake in all that and I have left my physics get rusty. Now, you have a template for other calorimetry or latent heat problems similar to this one.

1. A 0.40 kg iron horseshoe that is initially at 500 degrees Celsius is dropped into a bucket containing 20 kg of water at 22 degrees Celsius. What is the final equilibrium temperature? Neglect any energy transfer to or from the surrounding.

The iron loses as much heat as the water gains.

{\Delta}Q_{\text{iron}}=-{\Delta}Q_{\text{water}}

(mc{\Delta}T)_{\text{iron}}=-(mc{\Delta}T)_{\text{water}}

(1.5 \;\ kg)(448 \;\ J/kg\cdot C)(T-500C)=-(20 \;\ kg)(4186 \;\ J/kg\cdot C)(T-22)

Solve for T, the temperature in celsius.

thank you galactus ^^