Question

Suppose that 25 g of each of the following substances is initially at 29.0 ∘C. What is the final temperature of each substance upon absorbing 2.45 kJ of heat? Gold, silver, aluminum, and water.

Answer 1

The specific heat capacity for these materials are

Gold - 0.126 J/g ^oC

Silver - 0.233 J/g ^oC

Aluminium - 0.9 J/g ^oC

Water - 4.184 J/g ^oC

Q = m x Cp x dT Where m = mass and Cp = specific heats

For Gold

2450 J = 25 x 0.126 x dT

dT = 777.8 since initial temperature is 29 oC the final temperature is final - initial = 777.8

Final = 777.8 + 29^oC = 806.8 ^oC

For Silver

2450 J = 25 x 0.233 x dT

dT = 420.6 since initial temperature is 29 oC the final temperature is final - initial = 420.6

Final = 420.6 + 29^oC = 449.6 ^oC

For ALuminium

2450 J = 25 x 0.9 x dT

dT = 108.9 since initial temperature is 29 oC the final temperature is final - initial = 108.9

Final = 108.9 + 29^oC = 137.9 ^oC

For water

2450 J = 25 x 4.184 x dT

dT = 23.4 since initial temperature is 29 oC the final temperature is final - initial = 23.4

Final = 23.4 + 29^oC = 52.4 ^oC