Help Eric Solve 2 Heat Problems

[ericarnold86]

Thanks in advance for the help... I usually get Physics with ease, but for some reason these 2 threw me for a loop. If you have aol instant messanger, you could message me the answers at ericarnold86. Any help is greatly appreciated

#1:

A 0.018 kg cube of ice at 0.0 Celsius is added to 0.454 kg of soup at 87 Celsius. Assuming that the soup has the same specific heat capacity as water, find the temperature of the soup after the ice has melted. (There is a temperature change after the ice melts)


#2:

A 25.4 g ring that appears to be silver is heated to a temperature of 85 Celsius, and then is placed in a calorimeter containing .05 kg of water @ 24.8 Celsius. The calorimeter is not perfectly iinsulated, however, so that 0.11 kJ of energy is transferred by heat to the surroundings by the time a temperature of 25 celsius is reached. From the info provided, determine the specific heat capacity of the ring.

Thanks again

 

[ShawnD]

#1:

A 0.018 kg cube of ice at 0.0 Celsius is added to 0.454 kg of soup at 87 Celsius. Assuming that the soup has the same specific heat capacity as water, find the temperature of the soup after the ice has melted. (There is a temperature change after the ice melts)

I don't know what values you have for heat of fusion of water or heat capacity of water. I'm using 334000J/kg for heat of fusion and 4187J/kg for heat capacity of water.

energy in system does not change:
ice melt + water heat up + water cool down = 0
mLf + mc[tex] \Delta [/tex]T + mc[tex] \Delta [/tex]T = 0
(0.018)(334000) + (0.018)(4187)(Tf - 0) + (0.454)(4187)(Tf - 87) = 0

My calculator has a thing that solves for 0 = so I won't be showing any work.
Tf = 80.64 celcius

#2:

A 25.4 g ring that appears to be silver is heated to a temperature of 85 Celsius, and then is placed in a calorimeter containing .05 kg of water @ 24.8 Celsius. The calorimeter is not perfectly iinsulated, however, so that 0.11 kJ of energy is transferred by heat to the surroundings by the time a temperature of 25 celsius is reached. From the info provided, determine the specific heat capacity of the ring.

system energy does not change:
ring + water + surroundings = 0
mc[tex] \Delta [/tex]T + mc[tex] \Delta [/tex]T + 110J = 0
(0.0254)c(25 - 85) + (0.05)(4187)(25 - 24.8) + 110 = 0
c = 99.65 J/kg

 

[ericarnold86]

Those worked, thanks a lot