Question

In an effort to stay awake for an all-night study session, astudent makes a cup of coffee by first placing a 200 W electricimmersion heater in 0.300 kg of water.(a) How much heat must be added to the water to raiseits temperature from 20.0°C to 89.0°C? J(b) How much time is required? Assume that all of the heater powergoes into heating the water. s

Answer 1

Answer:

(a) Heat=86733 J =86.733 kJ

(b) Time=433.665 s =7.22775 minutes

Explanation:

Given data

$m_{water-mass}=0.300 kg\\ T_{1}=20.0^{o}C\\ T_{2}=89.0^{o}C$

To find

(a) Heat required

(b) Time required

Solution

For (a) heat required

By apply quantity heat equation

$Q_{water}=m*C_{water}*deltaT$

where C is specific heat and specific heat of water is 4190 J/kg.K

Substitute the given values

$Q_{water}=(0.300kg)*(4190J/kg.K)*(89.0-20.0)^{o}C\\Q_{water}=86733J\\or\\ Q_{water}=86.733kJ$

For (b) time required

From the definition of Power as "Power is the rate of energy to raise temperature"

$P_{power}=\frac{Heat}{Time}\\ Time=Q/P\\Time=\frac{86733J}{200J/s}\\ Time=433.665s\\or\\Time=7.22775 minutes$