Question A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.What is the specific heat of the substance?

Answer: 0.3832 Explanation: Givens m = 437.2 grams c = ? delta t = 69.8 – 19.3 H = 8460 J Formula H = m*c*Δt Solution 8460 = 437.2 * c * (69.8 – 19.3) 8460 = 437.2 * c * 50.5 8460 = 22078.6 * c c = 8460 / 22078,6 c = .3832 J / (oC * gr) Log in to Reply

Answer:0.3832

Explanation:Givensm = 437.2 grams

c = ?

delta t = 69.8 – 19.3

H = 8460 J

FormulaH = m*c*Δt

Solution8460 = 437.2 * c * (69.8 – 19.3)

8460 = 437.2 * c * 50.5

8460 = 22078.6 * c

c = 8460 / 22078,6

c = .3832 J / (oC * gr)