A sample of an unknown substance has a mass of 0 point 4 6 5 kilograms If 3,000.0J of heat is required to heat the substance from 50 point 0 times degree C to 100 point 0 times degree C , what is the specific heat of the substance? Use q is equal to m cap c sub p triangle cap t
Answer
A
0 point 1 2 9 the fraction with numerator cap j and denominator open paren g times degree cap c close paren
B
0 point 1 5 5 the fraction with numerator cap j and denominator open paren g times degree cap c close paren
C
0 point 0 0 7 7 5 the fraction with numerator cap j and denominator open paren g times degree cap c close paren
D
0 point 0 6 0 0 the fraction with numerator cap j and denominator open paren g times degree cap c close paren
Explanation
Use q = mcₚΔT, so cₚ = q/(mΔT). Convert 0.465 kg to 465 g, and calculate the temperature change: 100.0°C − 50.0°C = 50.0°C. Then cₚ = 3,000.0 J ÷ (465 g × 50.0°C) ≈ 0.129 J/(g·°C), which is choice A.
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