Well what are all the calculations you have
Anything carbon in composition
Q = mc(θ₂-θ₁)
47 calories = 10 g *c*(50.4 - 25)
47cal = 10*c* 25.4
47 /(10*25.4) = c
0.185 = c
Specific heat of iron = 0.185 cal/g°C
It should be a because the temperature and the atm are to low
