A312 g sample of a metal is heated to 257.128 °c and plunged into 200 g of water at a temperature of 41.933 °c. the final temperature of the water is 67.555 °c. assuming water has a specific heat capacity of 4.184 j/g °c, what is the specific heat capacity of the metal sample, in j/g °c)? assume no heat loss to the surroundings.

0.362 J/g °C

We are given:

Mass of metal = 312 g

Initial temperature of the metal = 257.128 °C

Mass of water = 200 g

Initial temperature = 41.933 °C

Final temperature of water = 67.555 °C

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

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

Step 1: Heat gained by water

Quantity of heat = mass × specific heat × change in temperature

Q=m × c × Δt

Change in temperature, Δt = 25.622 °C

Therefore,

Heat gained by water = 200 g × 4.184 J/g °C × 25.622 °C

                                    = 21,440.49 Joules

Change in temperature of the metal, Δt =(257.128 °C-67.555 °C)

                                                                 = 189.573 °C

Q = mcΔt

Assuming the specific heat capacity of the metal is c

Q= 312 g × 189.573 °C × c

 = 59,146.776c Joules

The heat released by the metal is equivalent to heat gained by water.

Therefore;

59,146.776c J = 21,440.49 J

 c = 21,440.49 J ÷  59,146.776

    = 0.362 J/g °C

Thus, the specific heat capacity of the metal is 0.362 J/g °C

explanation:

0.07 g/cm³

1.37 g/cm³