Question

A power station with an efficiency e generates W watts of electric power and dissipates D J of heat energy each second to the cooling water that flows through it. The cooling water increases its temperature by T degrees of Celsius. Find the mass of cooling water which flows through the plant each second. Specific heat capacity of water cw = 4,184 J/kgC e = 0.5 W = 4.2 x 108 watts D = 3.5 x 108 J T = 6 C Enter value in tons (1 ton = 1000 kg). Enter two digits after decimal point.

Answer 1

Answer: 13.94 tons/s

Explanation:

On adding heat energy to a substance, the temperature would be changed by a particular amount. This relationship between heat energy and temperature is often different for each material. The specific heat, is a value that describes how they relate.

Heat energy = mass flow rate * specific heat * Δ T

Q = MC (ΔΦ)

Heat energy, Q= 3.5*10^8J

Mass flow rate, M= ?

Specific heat, C= 4184j/KgC

Change in temperature, ΔΦ= 6°C

M = Q/CΔΦ

M = (3.5*10^8)/4184*6

M = 13942kg/s

M = 13.94 tons/s