A 312 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

Explanation:

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

Step 2: Heat released by the metal

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

Step 3: Calculate the specific heat capacity of the metal

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