Question

If 5,800 j of energy are applied to a 15.2 kg piece of lead, by how much does the temp change if the specific heat of lead is 0.128

Answer 1

Answer:

The answer will be 2.98K

Explanation:

Using the formula:

Q = mc∆T

Q= 5,800 (heat in joules)

m= convert 15.2kg to g which is 15200g (mass in grams)

c= 0.128 J/g °c (Specific heat capacity)

∆T=  what we need to find (temperature change)

5800J = 15200g x 0.128 x ∆T

= 2.98K