Answer :
Answer:
5183 sec.
Explanation:
- The power dissipated in the resistance, will be used to raise the temperature of the water completely.
- This power can be found applying Joule's law, which states the following:
[tex]P = I^{2}*R (1)[/tex]
- If the resistance of the heating coil can be assumed as constant with the temperature, we can find the current I applying Ohm's Law, as follows:
[tex]I = \frac{V}{R} = \frac{300V}{17 \Omega} = 17.7 A[/tex]
- Now, from (1) we can find P, as follows:
[tex]P = I^{2} * R = (17.7A)^{2} * 17 \Omega = 5294.1 J/s[/tex]
- The energy supplied by this power, is just the product of the power times the time during which the energy was delivered:
[tex]E = P* \Delta t (2)[/tex]
- This energy, will be supplied as heat to the mass of water, as stated by the following equation (assuming no heat losses out of the heater):
[tex]\Delta Q = c*m*\Delta T (3)[/tex]
- where c= specific heat of water = 4186 J/ºC*kg, m= 115 kg, and ΔT= 57ºC.
- From (2) and (3), if left sides are equal each other, so do right sides:
[tex]P* \Delta t = c*m*\Delta T[/tex]
- Replacing by the values, we can solve for Δt, as follows:
[tex]\Delta t = \frac{c*m*\Delta T}{P} = \frac{4186 J/deg*kg*115 kg*57 deg}{5294.1J/s} \\ \Delta t = 5183 sec.[/tex]
- The time required to raise the temperature of 115 kg of water, from 18ºC, to 75ºC, is 5183 sec (approximately 1hr 26').