A ball of thremal capacity `10 cal//^@C` is heated to the temperature of furnace it is then transferred into a vessel containing water. The water equivalent of vessel and the constents is 200 g. The temperature of the vessel and its contents rises from `10^@C` to `40^@C`. What is the temperature of furnace?

Options

A. `640^@C`
B. `64^@C`
C. `600^@C`
D. `100^@C`

Answer

A

Explanation

Thermal capacity of ball`=mc=10cal^(@)//C`
Let T bethe furnace temperature.
Water eq. of vessel and contents`=mc=200g`
Resultant temperature`=40^@C`.
According to principle of calorimetry.
Heat lost by hot body`=`heat gained by cold body.
`m_1c_1(T-40)=m_2c_2(40-10)`
`10(T-40)=200xx30`
`T=640^@C
`m_1xx1xx(50-30)=m_2xx1xx(80-10)`
`m_1xx20=m_2xx30`
or `(m_1)/(m_2)=(3)/(2)`
Mass of water from tank `A=(3)/(5)xx40=24kg`
Mass of water from tank `B=(2)/(5)xx40=16kg`