a bucket contains 8 kg of water at 25 degree Celsius and 2 kg of water at 80 degree celsius is poured into it neglecting the heat absorbed by the bucket calculate the final temperature of water
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Heat liberated by hot water = Heat absorbed by cold water
mSdT = m’SdT’
Divide both sides by S
mdT = m’dT’
2 kg × (80 - x) = 8 kg × (x - 25)
160 - 2x = 8x - 200
10x = 360
x = 36 °C
Final temperature of water is 36 °C
mSdT = m’SdT’
Divide both sides by S
mdT = m’dT’
2 kg × (80 - x) = 8 kg × (x - 25)
160 - 2x = 8x - 200
10x = 360
x = 36 °C
Final temperature of water is 36 °C
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