✨✨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.
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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
__________________________________
❤BE BRAINLY ❤
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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
__________________________________
❤BE BRAINLY ❤
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