Specific Heat Capacity Problem.(Please check.)

[shulien]

This is CALORIMETRY PROBLEM.

A 0.050kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15kg of water with an initial temperature of 21.0 C. The bolt and metal then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg*C, find the initial temperature of the metal.

the solution is:

Change in Temp of metal = (C of water *mass of water* Change in temp of water)/(C of metal*mass of metal)

Change of temp. of metal = [(4186 J/kg*C)(0.15kg)(25C-21C)]/[(899 j/kg*C)(0.050kg)
Change of temp. of metal = 55.87C

Change in temperature = Final Temperature-Initial Temperature
Initial Temperature = Final Temp.-Change in Temp
=25C-55.87C
= -30.87C


i want to ask why is the answer +80.87C from the book, instead of -30.87C?

i hope you could explain so i can also explain it to my students. is the sign of the answer has something to do with the HEAT REMOVAL and HEAT ABSORPTION?

Thanks!

 

[Doc Al]

the solution is:

Change in Temp of metal = (C of water *mass of water* Change in temp of water)/(C of metal*mass of metal)

This is off by a minus sign. The net change in thermal energy of "metal + water" is zero, so:
Change in Temp of metal = - (C of water *mass of water* Change in temp of water)/(C of metal*mass of metal)

Change of temp. of metal = [(4186 J/kg*C)(0.15kg)(25C-21C)/[(899 j/kg*C)(0.050kg)
Change of temp. of metal = 55.87C

The change in temp of the metal is -55.87C, not +. (The metal cools down.)

Change in temperature = Final Temperature-Initial Temperature
Initial Temperature = Final Temp.-Change in Temp
=25C-55.87C
= -30.87C

25C - (-55.87C) = +80.87C

i hope you could explain so i can also explain it to my students. is the sign of the answer has something to do with the HEAT REMOVAL and HEAT ABSORPTION?

Just keep in mind that the "heat" gained by the water must equal the "heat" removed from the metal. In terms of Q: For the water, Q = + (it heats up); for the metal, Q = - (it cools down). The net Q is zero.

 

[shulien]

This is off by a minus sign. The net change in thermal energy of "metal + water" is zero, so:
Change in Temp of metal = - (C of water *mass of water* Change in temp of water)/(C of metal*mass of metal)


The change in temp of the metal is -55.87C, not +. (The metal cools down.)


25C - (-55.87C) = +80.87C

Just keep in mind that the "heat" gained by the water must equal the "heat" removed from the metal. In terms of Q: For the water, Q = + (it heats up); for the metal, Q = - (it cools down). The net Q is zero.

thanks for the reply.