A metal ball of mass 0.1 kg is heated upto $500^{0} C$ and dropped into a vessel of heat capacity $800 J K^{- 1}$ and containing $0.5 k g$ water. The initial temperature of water and vessel is $30^{0} C$ . What is the approximate percentage increment in the temperature of the water? [Specific heat Capacities of water and metal are, respectively, $4200 J k g^{- 1} K^{- 1}$ and $400 J k g^{- 1} K^{- 1}$ ]

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 21.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Assume final temperature $= T^{0} C$
Heat less = Heat gain = $m s Δ T$
$\Rightarrow m_{B} s_{B} Δ T_{B} = m_{w} s_{w} Δ T_{w}$

$0.1 \times 400 \times (500 - T) = 0.5 \times 4200 \times (T - 30) + 800 (T - 30)$

$\Rightarrow 40 (500 - T) = (T - 30) (2100 + 800) \Rightarrow 2000 - 40 T = 2900 T - 30 \times 2900$

$\Rightarrow 2000 + 30 \times 2900 = T (2940) \Rightarrow T = {30.4}^{0} C \frac{Δ T}{T} \times 100 = \frac{6.4}{30} \times 100 = 21 %$

So the closest answer is 20%.

Watch 3-min video & get full concept clarity