<h3>
Answer:</h3>
0.362 J/g °C
<h3>
Explanation:</h3>
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
<h3>Step 2: Heat released by the metal </h3>
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
<h3>Step 3: Calculate the specific heat capacity of the metal</h3>
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
