Answer:
c=0.14J/gC
Explanation:
A.
2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.
B.
We can use the expression for heat transmission
[tex]Q=mc(T_2-T_1)[/tex]
In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say
[tex]Q_1=-Q_2[/tex]
for water we have to
c = 4.18J / g ° C
replacing we have
[tex]c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}[/tex]
I hope this is useful for you
A.
2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.
B.
Podemos usar la expresión para la transmisión de calor
[tex]Q=mc(T_2-T_1)[/tex]
En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir
[tex]Q_1=-Q_2[/tex]
para el agua tenemos que
c=4.18J/g°C
reemplazando tenemos
[tex]c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}[/tex]
