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Help!!! A student places 1.38 g of unknown metal at 99.68C into 60.50 g of water at 22.18C. The entire system reaches a uniform temperature at 31.68C. Calculate the specific heat of the metal.

Respuesta :

Kalahira
6.12
 The amount of heat lost by the piece of metal in changing temperature from 99.68 to 31.68 will exactly match the amount of heat gained by the water going from 22.18 to 31.68. So we can create an equality to demonstrate this and solve for the unknown specific heat. So 60.50*1*(31.68 - 22.18) = 1.38*X*(99.68 - 31.68) 60.50*1*9.5 = 1.38*X*68 574.75 = 93.84*X 6.124786871 = X Rounding to 3 significant digits gives a specific heat of 6.12
sebassandin

Answer:

[tex]C_{metal}=25.6\frac{J}{g^oC}[/tex]

Explanation:

Hello,

Thermal equilibrium is reached when two substances at different temperatures are allowed to stabilize thermodynamically until a constant temperature is reached, in this case 31.68 °C, in this manner, the suitable equation to model this phenomena is:

[tex]m_{metal}*C_{metal}*(T_{eq}-T_{metal})=-m_{water}*C_{water}*(T_{eq}-T_{water})[/tex]

Now, solving for the specific heat of the metal considering that the water's specific heat is [tex]4.18\frac{J}{g^oC}[/tex] and the given data, one obtains:

[tex]C_{metal}=\frac{-m_{water}*C_{water}*(T_{eq}-T_{water})}{m_{metal}*(T_{eq}-T_{metal})} \\C_{metal}=\frac{-(60.50g)*(4.18\frac{J}{g^oC} )*(31.68^oC-22.18^oC)}{(1.38g)*(31.68^oC-99.68^oC)} \\\\C_{metal}=25.6\frac{J}{g^oC}[/tex]

Best regards.

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