Answer:
A) 50.0 g Al
Explanation:
We can calculate the temperature change of each substance by using the equation:
[tex]\Delta T=\frac{Q}{mC_s}[/tex]
where
Q = 200.0 J is the heat provided to the substance
m is the mass of the substance
[tex]C_s[/tex] is the specific heat of the substance
Let's apply the formula for each substance:
A) m = 50.0 g, Cs = 0.903 J/g°C
[tex]\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C[/tex]
B) m = 50.0 g, Cs = 0.385 J/g°C
[tex]\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C[/tex]
C) m = 25.0 g, Cs = 0.79 J/g°C
[tex]\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C[/tex]
D) m = 25.0 g, Cs = 0.128 J/g°C
[tex]\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C[/tex]
E) m = 25.0 g, Cs = 0.235 J/g°C
[tex]\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C[/tex]
As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.
