Answer:
The temperature after half an hour is [tex]19.3002^{\circ}[/tex]
Solution:
As per the question;
Metabolic rate of the person is 3.0105 J/h
Temperature, T = [tex]19.30^{\circ}[/tex]
Mass of the water, [tex]m_{w} = 1.2103 kg[/tex]
Time duration, t = 0.5 h = 30 min = 180 s
Now,
Heat, [tex]Q = ms\Delta t[/tex]
Thus heat transfer in half an hour:
[tex]Q = 3.0105\times 0.5 = 1.505 J[/tex]
Now, the temperature of water after half an hour, T' is given by:
[tex]Q = m_{w}s\Delta T = ms(T' - T)[/tex]
where
s = 4186 J
[tex]1.505 = 1.21103\times 4186\times (T' - 19.30^{\circ})[/tex]
[tex]T' = 19.3002 ^{\circ}[/tex]
