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5 grams of ice at 0 degree C.is mixed with 10 grams of water at 50 degrees C.then resultant temperature is

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specific heat of water = 4.186 J/g degreeC

First 5g ice melts and then its temperature increases by gaining heat from 10g water. let the resultant temperature is T.

Energy gained by 5g ice = energy lost by 10g water

5*(336) + 5*4.186*(T-0) = 10*4.186*(50-T)

1680 + 20.93T = 2093 - 41.86T

(20.93+41.86)T = 2093-1680

62.79T = 413

T = 413/62.79 =

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Resultant temperature is T.

Ice could melt and become water. Water at 50° C will cool down.

Heat lost by Hot water = Heat gained by ice

= heat energy for melting of 5g of ice

+ heat energy for increase of temperature of 5g water from 0°C to T°C

m1 s1 (T2 - T) = m2 * L + m2 s2 (T - T1)

10g * 4.181J/g/°K * (50 - T) °K = 5g * 334 J/g + 5g * 4.181J/g/°K * (T - 0) °K

or, 10 * 4.181 (50-T) = 5*334 + 5 * 4.181 (T-0)

41.81 (50-T) = 1,670 + 20.905 T

T = 420.5 / 62.715 = 6.705° C

Ice could melt and become water. Water at 50° C will cool down.

Heat lost by Hot water = Heat gained by ice

= heat energy for melting of 5g of ice

+ heat energy for increase of temperature of 5g water from 0°C to T°C

m1 s1 (T2 - T) = m2 * L + m2 s2 (T - T1)

10g * 4.181J/g/°K * (50 - T) °K = 5g * 334 J/g + 5g * 4.181J/g/°K * (T - 0) °K

or, 10 * 4.181 (50-T) = 5*334 + 5 * 4.181 (T-0)

41.81 (50-T) = 1,670 + 20.905 T

T = 420.5 / 62.715 = 6.705° C