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
![Q=mc(T_2-T_1)](https://tex.z-dn.net/?f=Q%3Dmc%28T_2-T_1%29)
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
![Q_1=-Q_2](https://tex.z-dn.net/?f=Q_1%3D-Q_2)
for water we have to
c = 4.18J / g ° C
replacing we have
![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}](https://tex.z-dn.net/?f=c_%7Bmetal%7D%2A%28500g%29%28100%5C%C2%B0C-25%5C%C2%B0C%29%3D-%28250g%29%284.18%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%29%2820%5C%C2%B0C-25%5C%C2%B0C%29%5C%5Cc_%7Bmetal%7D%3D0.14%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D)
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
![Q=mc(T_2-T_1)](https://tex.z-dn.net/?f=Q%3Dmc%28T_2-T_1%29)
En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir
![Q_1=-Q_2](https://tex.z-dn.net/?f=Q_1%3D-Q_2)
para el agua tenemos que
c=4.18J/g°C
reemplazando tenemos
![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}](https://tex.z-dn.net/?f=c_%7Bmetal%7D%2A%28500g%29%28100%5C%C2%B0C-25%5C%C2%B0C%29%3D-%28250g%29%284.18%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%29%2820%5C%C2%B0C-25%5C%C2%B0C%29%5C%5Cc_%7Bmetal%7D%3D0.14%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D)
Answer:
350 N/m
Explanation:
If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m
k = 70 / 0.2 = 350 N/m
The spring constant is 350 N/m
Answer:-2.86*10⁻⁴
Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B
Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴
ΔP = P₂-P₁ ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷