A sample 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