Answer:
Approximately \(31\, 350\; {\rm s}\) (approximately \(8.71\) hours.)
Explanation:
Approach this question in the following steps:
- Find the mass of the water that needs to be heated.
- Using the specific heat capacity of water, find the amount of energy that is required.
- To find the time required, divide the amount of energy required by the rate at which the water absorbs the energy.
The mass of water that needs to be heated is:
\(\begin{aligned}m &= \rho \, V = (150\; {\rm L})\, \left(\frac{1\; {\rm kg}}{1\; {\rm L}}\right) = 150\; {\rm kg}\end{aligned}\).
Assuming that no phase change such as melting or vaporization happens, the amount of energy required to raise the temperature of an object is:
\(Q = c\, m\, \Delta T\),
Where:
- \(c\) is the specific heat capacity of the material,
- \(m\) is the mass of the object, and
- \(\Delta T\) is the change in temperature.
In this question:
- \(c \approx 4.18\times 10^{3}\; {\rm J\cdot kg \cdot K^{-1}}\),
- \(m = 150\; {\rm kg}\), and
- \(\Delta T = (70.0 - 10.0) = 60.0\; {\rm K}\).
Hence, the amount of energy that the water needs to absorb would be:
\(\begin{aligned}Q &= c\, m\, \Delta T \\ &= (4.18 \times 10^{3}\; {\rm J\cdot kg \cdot K^{-1}})\, (150\; {\rm kg})\, (60.0\; {\rm K}) \\ &\approx 3.76 \times 10^{7}\; {\rm J} \end{aligned}\).
It is given that the rate of energy input to the water is \(20.00\%\) that of the energy rating of this heater, \(6.00\; {\rm kW}\). In other words, the rate of useful energy input to the water would be:
\(\begin{aligned} & (20.00\%)\, (6.00\; {\rm kW}) \\ =\; & (20.00\%)\, (6.00 \times 10^{3}\; {\rm W}) \\ =\; & 1.2 \times 10^{3}\; {\rm W}\end{aligned}\).
To find the time required, divide the energy input by the rate of energy delivery:
\(\begin{aligned}t &\approx \frac{3.76 \times 10^{7}\; {\rm J}}{1.2 \times 10^{3}\; {\rm W}} &\approx 31\, 350\; {\rm s}\end{aligned}\).
