Question

Water (3070 g ) is heated until it just begins to boil. if the water absorbs 5.03×105 j of heat in the process, what was the initial temperature of the water?

Answer 1

Answer: $60.8\°C$  

Explanation:

The heat absorbed $Q$ is given by:

$Q=(m) (c) \Delta T$  (1)  

Where:  

$Q=5.03 (10)^{5} J$  

$m=3070 g$  is the mass of water

$c=4.18 \frac{J}{g \°C}$  is the specific heat of  water  

$\Delta T=T_{f}-T_{i}$  is the variation in temperature, being  the final temperature $T_{f}=100\°C$ which is the boiling temperature of water  

$T_{i}$ is the initial temperature

Finding $\Delta T$:

$\Delta T=\frac{Q}{(m)(c)}$   (2)  

$\Delta T=\frac{5.03 (10)^{5} J}{(3070 g)(4.18 \frac{J}{g \°C})}$   (3)  

$\Delta T= 39.19 \°C$  (4)  

Knowing $\Delta T$  and $T_{f}$ we can find $T_{i}$:

$\Delta T=T_{f}-T_{i}$  

$T_{i}=T_{f} - \Delta T$ (5)

$T_{i}=100 \°C - 39.19 \°C$ (6)

Finally:

$T_{i}=60.8 \°C$ This is the initial temperature of the water