Question

The burner on an electric stove has a power output of 2.0 kW. A 650 g stainless steel tea kettle is filled with 20∘C water and placed on the already hot burner. If it takes 2.70 min for the water to reach a boil, what volume of water, in cm^3, was in the kettle?

Answer 1

The value is
`V  =  900 \  cm^3`

Step-by-step explanation:

From the question we are told that

The power output is
`P_(out) =  2.0 \  kW  =  2.0 *10^(3) \  W`

The mass of the steel is
`m=  650 \  g  =  (650)/(1000)  =  0.650 \ kg`

The temperature of the water is
`T  =  20^o C`

The time take is
`t  =  2.70 \  minutes =  2.70 *60  = 162 \  s`

Generally the quantity of heat energy given out by the electric stove is mathematically represented as

`Q =  P * t`

=>
`Q =  2.0 *10^(3)  * 162`

=>
`Q =  324000 \ J`

This energy can also be mathematically represented as

`Q =   \Delta T  * m  c_s *  +  m_w  * c_w *  \Delta T`

Here
`c_s` is the specific heat of stainless steel with value
`c_s =  450\ J/C/kg`

tex]c_s[/tex] is the specific heat of water with value
`c_s =  4180\ J/C/kg`

m_w is the mass of water which is mathematically represented as

`m_w  =  \rho_w  *  V`

=>
`m_w  =  1000  *  V`

So

`324000 =   (100 -20 )  * 0.650  * 450 *  + 1000V * 4180 *  (100-20)`

`V  =  0.0008989 \  m^3`

converting to
`cm^3`

`V  =  900 \  cm^3`