Respuesta :

[tex]\begin{gathered} \text{Question 5)} \\ Q=52,000J \\ m=2kg \\ C=130\text{ J/Kg K} \\ \Delta T=\text{?} \\ Q=Cm\Delta T \\ \text{Solving }\Delta T \\ \Delta T=\frac{Q}{Cm} \\ \Delta T=\frac{52,000J}{(130\text{ J/Kg K)(}2kg\text{)}} \\ \Delta T=200K,\text{ but Kelving and }degree\text{ Celsius have the same change} \\ \text{Hence} \\ \Delta T=200\text{ \degree{}C} \\ \\ \text{Question 6} \\ T1=350\text{ K} \\ \Delta T=200\text{ K} \\ T2=\text{?} \\ \Delta T=T2-T1 \\ \text{Sol}v\text{ing T2} \\ T2=\Delta T+T1 \\ T2=200\text{ K}+350\text{ K} \\ T2=550\text{ K} \\ \text{The final temperature is }550\text{ K} \end{gathered}[/tex]

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