Final answer:
To find the heat released by 30 g of water cooling from 96°C to 25°C, the specific heat capacity formula q = m × c × ΔT is used, resulting in a heat release of 2130 calories.
Step-by-step explanation:
To calculate how much heat is released when 30 g of water at 96°C cools to 25°C, we can use the specific heat equation. The specific heat of water is given as 1 cal g-1°C-1.
The equation for calculating the heat (q) released or absorbed during a temperature change is:
q = m × c × ΔT
where:
- m = mass of the water (30 g)
- c = specific heat capacity of water (1 cal g-1°C-1)
- ΔT = change in temperature (96°C - 25°C = 71°C)
Substituting the values into the equation:
q = 30 g × 1 cal g-1°C-1 × 71°C
q = 30 × 71
q = 2130 calories
Therefore, 2130 calories of heat are released when 30 g of water cools from 96°C to 25°C.