Question

1. suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20°c. what would have been the final temperature of the water?

Explanation:

Step1: Convert units

Convert energy \( Q = 65\space kJ=65000\space J \), mass \( m = 450\space g \), specific - heat capacity of water \( c = 4.18\space J/(g\cdot^{\circ}C) \), initial temperature \( T_1=20^{\circ}C \). Use the heat - transfer formula \( Q = mc\Delta T=mc(T_2 - T_1) \).

Step2: Solve for \( T_2 \)

Rearrange the formula \( T_2=\frac{Q}{mc}+T_1 \).
Substitute the values: \( T_2=\frac{65000}{450\times4.18}+20 \).
First, calculate \( \frac{65000}{450\times4.18}=\frac{65000}{1881}\approx34.56 \).
Then \( T_2 = 34.56+20=54.56^{\circ}C \approx55^{\circ}C \).

Answer:

\( 55^{\circ}C \)