Answer:
Approximately [tex]31\, 350\; {\rm s}[/tex] (approximately [tex]8.71[/tex] hours.)
Explanation:
Approach this question in the following steps:
The mass of water that needs to be heated is:
[tex]\begin{aligned}m &= \rho \, V = (150\; {\rm L})\, \left(\frac{1\; {\rm kg}}{1\; {\rm L}}\right) = 150\; {\rm kg}\end{aligned}[/tex].
Assuming that no phase change such as melting or vaporization happens, the amount of energy required to raise the temperature of an object is:
[tex]Q = c\, m\, \Delta T[/tex],
Where:
In this question:
Hence, the amount of energy that the water needs to absorb would be:
[tex]\begin{aligned}Q &= c\, m\, \Delta T \\ &= (4.18 \times 10^{3}\; {\rm J\cdot kg \cdot K^{-1}})\, (150\; {\rm kg})\, (60.0\; {\rm K}) \\ &\approx 3.76 \times 10^{7}\; {\rm J} \end{aligned}[/tex].
It is given that the rate of energy input to the water is [tex]20.00\%[/tex] that of the energy rating of this heater, [tex]6.00\; {\rm kW}[/tex]. In other words, the rate of useful energy input to the water would be:
[tex]\begin{aligned} & (20.00\%)\, (6.00\; {\rm kW}) \\ =\; & (20.00\%)\, (6.00 \times 10^{3}\; {\rm W}) \\ =\; & 1.2 \times 10^{3}\; {\rm W}\end{aligned}[/tex].
To find the time required, divide the energy input by the rate of energy delivery:
[tex]\begin{aligned}t &\approx \frac{3.76 \times 10^{7}\; {\rm J}}{1.2 \times 10^{3}\; {\rm W}} &\approx 31\, 350\; {\rm s}\end{aligned}[/tex].
