When running at a constant temperature, one practical engine goes through 2400
cycles every minute. In one complete cycle of this engine, 114 J of energy has to be
removed by a coolant so that the engine runs at a constant temperature. The
temperature of the coolant rises by 18 °C as it passes through the engine.
Calculate the volume of the coolant that flows through the engine in one second.
specific heat capacity of coolant = 3.8 × 10^3 J kg–1 K–1
density of coolant = 1.1 × 103 kg m–3
Can someone please help me with this question I can't seem to get the same answer as the one in the mark scheme
-so first I found Q= (2400/60)x 114 = 4560J
-then I used Q=mcT with T=18+273 to find the mass
-then I used density=m/v to rearrange for volume and I got 3.8x10^-6 and the answer is meant to be = 6.06 × 10^−5