Specific Heat and Calorimetry

Question

The temperature of an object increases by 71.0 °C when it absorbs 3691 J of heat. Calculate the heat capacity of the object.

C= Answer J/°C

The mass of the object is 403 g. Use the table of specific heat values to identify the composition of the object.

SubstanceSpecific heat (J/(g⋅°C))
gold	0.129
silver	0.235
iron	0.444
aluminum	0.900

The object is composed of gold, silver, iron, or aluminum?

Answer 1

Using the formula Q=mc*delta T, where Q = heat, m= mass, c = specific heat,and delta T = the change in temperature;

The problem gives the following information:

Q = 3691 J

m = 403 g

delta T = 71.0 ºC

specific heat = unknown

Q=m×c×delta T

Rearrange our formula to solve for our unknown quantity.

c=Q÷(m×delta T)

c=3691 J ÷ (403 g × 71.0 ºC)

c=0.129 J/(g⋅°C) our substance is Gold