Answer:
The new temperature of the water bath 32.0°C.
Explanation:
Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)
Initial temperature of the water = [tex]T_1=33.9^oC=33.9+273K=306.9 K[/tex]
Final temperature of the water = [tex]T_2[/tex]
Specific heat capacity of water under these conditions = c = 4.18 J/gK
Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J
( 1kJ=1000 J)
[tex]Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)[/tex]
[tex]-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)[/tex]
[tex]-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)[/tex]
[tex]T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC[/tex]
The new temperature of the water bath 32.0°C.
