Asample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 100.0°c, what is the specific heat of the substance? use mc008-1.jpg. 0.00775 j/(gi°c) 0.0600 j/(gi°c) 0.129 j/(gi°c) 0.155 j/(gi°c)

Answer : Option C) 0.129 J/g°C

Explanation : The given information has about Q which is heat energy (3,000 J);

m is the mass of a substance (0.465 kg);

∆T is the measure for the change in temperature (100 - 50) = 50°C;

To find the specific heat of the given unknown substance, we have to use the formula as;

q = m x C x ∆T; where c is the specific heat;

So, substituting the above values in this equation we get;

C = q/ (m x ∆T)

Therefore, it will be C = 3000 / (0.465 X 50) = 129.0 J/Kg°C

But we want the answer in grams, not in kilograms;

So, it will be 129.0 / 1000 = 0.129 J/g°C

Therefore, the specific heat of the unknown substance is 0.129 J/g°C