If 100 kilojoules of energy is used to heat 500 grams of water what is the temperature change of the water

47.8 °C

Explanation:

Use the heat equation:

q = mCΔT

where q is the heat absorbed/lost,

m is the mass of water,

C is the specific heat capacity,

and ΔT is the change in temperature.

Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.

100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT

ΔT = 47.8 °C

need to see circuit to get time constant