Question

✨✨A hot iron ball of mass 0.2 kg is dropped into 0.5 kg of water at 10°C.The resulting temperature is 30°C. calculate the temperature of hot ball. Specific heat capacity of iron = 336 J kg^-1 k^-1 and Specific heat capacity of water = 4.2 * 10³ J kg^-1 K^-1.

Answer 1

ANSWERS :-


Let temperature of hot ball be t°C

fall in temperature of Ball = ( t - 30°C )

Rise in temperature of water = ( 30 - 10 )°C

Heat energy given by ball = 0.2 * 336 * (t - 30) J

Heat energy taken by water

= 0.5 * ( 4.2 * 10³) * ( 30 - 10 ) J

Assuming that there is no loss of heat energy,

0.2 * 336 * ( t - 30 ) = 0.5 * 4.2 * 10³ * 20

or,

67.2 t – 2016 = 42000

or,

67.2 t = 42000 + 2016

or,

67.2 t = 44016

or,

t = 44016/67.2

t = 655°C

Thus,

temperature of hot ball was 655 °C

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Answer 2

Answer:

here is your answer............