Question

750.0g of water that was just boiled (heated to 100.0/c loses 78.45KJ of heat as it cools.What is the final temperature of water ?

Answer 1

• Mass=m=750g=0.75kg
• Q=78.45\times 10^3J=78450J
• T_i=100°C
• T_f=?
• Specific heat capacity=c=4200J/kg°C

According to thermodynamics

$\\ \tt\bull\rightarrowtail Q=mc\Delta T$

$\\ \tt\bull\rightarrowtail Q/mc=\Delta T$

$\\ \tt\bull\rightarrowtail \Delta T=\dfrac{78450}{0.75(4200)}$

$\\ \tt\bull\rightarrowtail \Delta T=24.9°C$

$\\ \tt\bull\rightarrowtail T_i-T_f=24.9$

$\\ \tt\bull\rightarrowtail T_f=100-24.9=76.1°C$