Acup holding 125.12g of water has an initial temperature of 26.8 degrees c. after a 35.08g piece of metal, at 99.5 degrees c, is transferred to the water, the temperature rises to 29.7 degrees c. what is the specific heat of the metal?

0.620 J/g°C

Heat gained or absorbed, Q by a substance is calculated by;

Q = mass × specific heat capacity × Change in temperature

In this case we are given;

We are required to calculate the specific heat capacity of the metal;

Specific heat capacity of water = 4.184 J/g°C

Temperature change of water = 29.7 °C - 26.8°C

                                                   = 2.9 °C

But, Q = m×c×ΔT

Thus, Heat = 125.12 g × 2.9°C × 4.184 J/g°C

                  = 1518.156 Joules

Specific heat capacity of the metal = x J/g°C

Temperature change of the metal = 29.7 °C - 99.5°C

                                                         = -69.8 °C

But, Q = mcΔT

Therefore;

Heat lost by the meatl = 35.08 g × x J/g°C × 69.8 °C

                                     = 2.448.584x Joules

The heat gained by water is equal to the heat lost by the metal

Therefore;

1518.156 Joules = 2.448.584x Joules

x = 1518.156 J ÷ 2.448.584 J

   = 0.620 J/g°C

Therefore, the specific heat of the metal is 0.620 J/g°C

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

answeridentity;