Answer:
See explanation
Step-by-step explanation:
In this case, we can start with the calculation of the total energy released by the 20 matches. One match release 838.2 J of energy ( 1 match = 838.2 J). So:
With this in mind, we have to find al the conversion ratios to "Jolues", so:
![1~Cal=4.184~J](https://img.qammunity.org/2021/formulas/chemistry/college/tv1qo6f1wwcrfa7s03e3s6aul06mr80ozt.png)
![1~KCal=4184~J](https://img.qammunity.org/2021/formulas/chemistry/college/4gwua1p659udfm95dse4i5mppvpg337kjt.png)
![0.000239~food~calories=1~J](https://img.qammunity.org/2021/formulas/chemistry/college/wzprivk8zwqgnudzz2bfyws4jie4r65lpu.png)
Now, we can do the conversions:
![16764~J(1~KJ)/(1000~J)=16.76~KJ](https://img.qammunity.org/2021/formulas/chemistry/college/sdrr6l6rlzjlvhjerqhl9xb0iztoi6uk79.png)
![16764~J(1~Cal)/(4.184~J)=4.005x10^3~Cal](https://img.qammunity.org/2021/formulas/chemistry/college/97mar0ckps4mojhbe04z0zgsegycg8wz65.png)
![16764~J(1~KCal)/(4184~J)=4.007~KCal](https://img.qammunity.org/2021/formulas/chemistry/college/t895fcs2x9koiqaym1laca0whj9ohjw0y0.png)
![16764~J(0.000239~food calories)/(1~J)=4.006~food calories](https://img.qammunity.org/2021/formulas/chemistry/college/6e3904zryxarzebnn8933lsmhru30v0f5l.png)
I hope it helps!