GWCB002 Question 3

Explanation

Part (1)

The amount of heat (energy) $Q\ \mathrm{J}$Q J required to raise the temperature of a substance with mass $m\ \mathrm{g}$m g by $\Delta T\ \mathrm{K}$ΔT K is expressed by the following equation, assuming the specific heat capacity is $c\ \mathrm{J}/(\mathrm{g}\cdot\mathrm{K})$c J/(g⋅K).

$Q = m c \Delta T$Q=mcΔT

The change in temperature $\Delta T$ΔT of the hot water in this problem is calculated as follows:

$\Delta T = T_1 - T_0 = 333.15 - 288.15 = 45.00\ \mathrm{K}$ΔT=T1​−T0​=333.15−288.15=45.00 K

Since the thermal efficiency $\eta_0 = 1.00$η0​=1.00 ($100\%$100%), the total energy $A_1$A1​ generated by the heating device and used to heat the water is equal to the heat absorbed by the water. Substituting the values from the constraints, we can calculate this as follows:

$A_1 = 2250 \times 4.184 \times 45.00 = 423630\ \mathrm{J}$A1​=2250×4.184×45.00=423630 J

Subquestion (2)

The average power consumption $P\ \mathrm{W}$P W is obtained by dividing the total energy consumed $W\ \mathrm{J}$W J by the time $t\ \mathrm{s}$t s taken for heating.

$P = \frac{W}{t}$P=tW​

According to the problem statement, the time taken for heating is “4 minutes and 30 seconds.” Convert this to seconds.

$t = 4 \times 60 + 30 = 270\ \mathrm{s}$t=4×60+30=270 s

Substituting the total energy $A_1 = 423630\ \mathrm{J}$A1​=423630 J found in Subquestion (1) and the time $t = 270\ \mathrm{s}$t=270 s, the average power consumption $A_2$A2​ can be calculated as follows.

$A_2 = \frac{423630}{270} = 1569\ \mathrm{W}$A2​=270423630​=1569 W

Calculation of the Input Format

The value to be found is the sum of the answer $A_1$A1​ from Part (1) and the answer $A_2$A2​ from Part (2).

$A_1 + A_2 = 423630 + 1569 = 425199$A1​+A2​=423630+1569=425199

The final natural number to be entered is 425199.