When you take a bath, how many kilograms of hot water (48°C) must you mix with cold water (10°C) so that the temperature of the bath is 36°C? The total mass of water (hot plus cold) is 192 kg. Ignore any heat flow between the water and its external surroundings.
2007-12-08 15:19:16 · 3 answers · asked by koetjet24 1 in Science & Mathematics ➔ Physics
Q = m c ΔT
Heat gained by cool water = heat lost by hot water
( m_c ) ( c ) ( 36 - 10 ) = ( m_h ) ( c ) ( 36 - 48 )
m_c + m_h = 192
m_c = mass of cold water
m_h = mass of hot water
c = specific heat of water
So substitute m_c = 192 - m_h into the first equation and solve for m_h.
2007-12-08 15:28:45 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
Consider a mass of water "m" and "n" where m is the hot water and n is the cold water. Heat lost by one must be gained by another. In equating them their specific heat is the same and thus unnecessary in the calculations. From this the relationship is:
m*(36°C-48°C) = -n(36°C-10°C)
The ratio of temperature changes is related to the ratio of masses.
-m/n = -26°C/12°C
From this mass ratio and knowing the total mass is 192 kg it follows that:
m = (26/38)192kg = 131 kg
n = (12/38)192kg = 60.6 kg
2007-12-08 23:35:12 · answer #2 · answered by bloodninja 3 · 0⤊ 0⤋
Heat loss = - heat gained
let:
m - be the weight of the hot water
192 - m = be the cold water
m ( 48 - 36) = - (192 - m)(10-36)
12m = - (192 - m) (-26)
m = (192 x 26) / (26 +12)
m = 131.37 Kg
2007-12-08 23:39:16 · answer #3 · answered by bernie_bph 5 · 0⤊ 0⤋