This one is beta decay (the -1 subscript tells us that)
between two oppositely charged ions
Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C
