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The temperature of 170 g of a material decreases by 20.0⁰C when it loses 3050 J of heat. What is its specific heat

Sagot :

jbferrado

Answer:

0.897 J/g.⁰C

Explanation:

Given the following data:

m = 170 g (mass)

ΔT = 20.0⁰C (change in temperature)

q = 3050 J (amount of heat)

The amount of heat (q) is calculated as follows:

q = m x Cp x ΔT

Thus, we introduce the data in the mathematical expression to calculate the specific heat (Cp):

Cp = q/(m x ΔT) = 3050 J/(170 g x 20.0⁰C) = 0.897 J/g.⁰C

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