<h3>
Answer:</h3>
0.620 J/g°C
<h3>
Explanation:</h3>
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;
- Mass of water = 125.12 g
- Initial temperature of water = 26.8 °C
- Initial temperature of the metal = 99.5°C
- Final temperature of the mixture = 29.7 °C
We are required to calculate the specific heat capacity of the metal;
<h3>Step 1 : Heat absorbed by water </h3>
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
<h3>Step 2; Heat lost by the metal </h3>
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
<h3>Step 3: C;aculating the specific heat capacity of the metal </h3>
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
