Specific Heat Capacity of a metal bar placed into water

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Homework Help Overview

The discussion revolves around the specific heat capacity of a metal bar placed into water, involving thermal energy transfer and temperature changes in both substances.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to relate the thermal energy gained by water to the cooling of the metal, questioning how these values interact in the context of specific heat capacity calculations.

Discussion Status

There is an ongoing exploration of how to set up the equations for the specific heat of the metal, with some participants suggesting the need to define an unknown for the metal's specific heat and questioning the relevance of previously calculated values.

Contextual Notes

Participants are navigating the constraints of using specific heat capacities for both water and metal, with uncertainty about how to incorporate these into their calculations.

lxhull
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Homework Statement
A thermos bottle contains 0.150 kg of water at 4.1 °C. When 9.00 x 10^-2 kg of a metal, initially at 96.2 °C, is put into the water, the temperature of the water rises to 21.7 °C. Calculate the specific heat of the metal
Relevant Equations
C= Eth/mT
Previously solved thermal energy gained by water as
Eth= 0.15(4180)(17.6) = 11035.2 J
Not sure if its relevant
 
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lxhull said:
Not sure if its relevant
It is. How does it relate to the cooling of the metal?
 
haruspex said:
It is. How does it relate to the cooling of the metal?
That's the problem, I don't know. It seems like it can't be part of the equation for the metal's shc because it used the waters shc, so I can't figure it out.
 
lxhull said:
That's the problem, I don't know. It seems like it can't be part of the equation for the metal's shc because it used the waters shc, so I can't figure it out.
Just write the corresponding equation for the metal's change in temperature. Create an unknown for the metal's s.h.
 
lxhull said:
Homework Statement:: A thermos bottle contains 0.150 kg of water at 4.1 °C. When 9.00 x 10^-2 kg of a metal, initially at 96.2 °C, is put into the water, the temperature of the water rises to 21.7 °C. Calculate the specific heat of the metal
Relevant Equations:: C= Eth/mT

Previously solved thermal energy gained by water as
Eth= 0.15(4180)(17.6) = 11035.2 J
Not sure if its relevant
Corrrct so far.
 

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