Question

A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of heat is lost from the water. The temperature rise of water is : (a) 21.50 (b) 12.0C (c)31.1C (d) 50.0C

Answer 1

ΔT= 11.94 °C

Step-by-step explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400 KJ

Heat input to the water = 1 KW

Heat input the water= 1 x 15 x 60

=900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

`C_p=4.187 (KJ)/(kg-K)`

`Q=mC_p\Delta T`

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So rise in temperature of water ΔT= 11.94 °C