1.) What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 10000g of the substance increases in temperature from 283 K to 303 K?

Formula: Q = mc∆T Mass = 10000g C = 2500 J/kg x K Ti = 283 K Tf = 303 K

10000g ---> 10 kg

Solution: (10 kg)(2500 J/kg x K)(303 K - 283 K) = 50000 J/kg x K

2.) A 0.400kg brick of copper is heated from 295 K to 325 K, how much heat is absorbed by the iron?

Mass: 0.400kg C= 450J/kg x K Ti = 295 K Tf = 325 K

Solution: Q = mc∆T = (0.400kg)(450 J/kg x K)(325 K – 295 K) = Q = 5.40 x 103 J

1.) What is the specific heat of a substance that absorbs 2500joules of heat when a sample of 10000g of the substance increases in temperature from 283 K to 303 K?Formula: Q = mc∆T

Mass = 10000g

C = 2500 J/kg x K

Ti = 283 K

Tf = 303 K

10000g ---> 10 kg

Solution:

(10 kg)(2500 J/kg x K)(303 K - 283 K) = 50000 J/kg x K

2.) A 0.400kg brick of copper is heated from 295 K to 325 K, how much heat is absorbed by the iron?Mass: 0.400kg

C= 450J/kg x K

Ti = 295 K

Tf = 325 K

Solution: Q = mc∆T

= (0.400kg)(450 J/kg x K)(325 K – 295 K) =

Q = 5.40 x 103 J