You have 750 grams of water at 80° celsius. which of the following would lower the temperature of the water by 10° celsius? (1 point)
adding 750 grams of water at 50° celsius
adding 325 grams of water at 60° celsius
adding 750 grams of water at 60° celsius
adding 1000 grams of water at 55° celsius

adding 750 grams of water at 60° Celsius .

Explanation:

Q = m.c.ΔT,

Where, Q is the amount of heat lost by water (Q = ??? J).

m is the mass of water (m = 750.0 g).

c is the specific heat capacity of the water (c = 4.18 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.

Now, we can calculate the Q that is gained by the different added amounts of water:

ΔT = 70.0°C - 50.0°C = 20.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.

ΔT = 70.0°C - 55.0°C = 15.0°C,

∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.

adding 750 grams of water at 60° Celsius

5.6⋅10−^19  j

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