Question

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

Answer 1

adding 750 grams of water at 60° Celsius .

Step-by-step explanation:

• We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:

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:

• adding 750 grams of water at 50° Celsius :

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.

• adding 325 grams of water at 60° Celsius :

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.

• adding 750 grams of water at 60° Celsius :

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.

• adding 1000 grams of water at 55° Celsius:

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.

• So, the right choice is:

adding 750 grams of water at 60° Celsius