Physics practice problems help?

Question

1. Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C. The specific heat of the solid:

   a. is equal to that of the fluid. b. is less than that of the fluid. c. is more than that of the fluid. 

2.  

   1. A puddle holds 150 g of water. If 0.50 g of water evaporates from the surface, what is the approximate temperature change of the remaining water? (Lv = 540 cal/g)

      a. +1.8 C° b. –1.8 C° c. +0.18 C° d.

Answer 1

(1)

Temperature of the solid Ts = 70°C

 Mass of the solid m_s= 20 g

 Specific heat  of the solid= Cs

 Temperature  of the fluid Tf = 20°C

 Mass of the fliud m_f = 100 g

 Specific heat of the fluid= Cf

Temperature at thermal equilibrium  Te = 30°C

ΔTs = Te - Ts  = 30 - 70 = - 40

ΔTf = Te - Tf  = 30 - 20 = 10

At thermal equilibrium,Qhot = -Qcold

m_fCf (ΔTf) = -m_sCs(ΔTs)

100  *Cf*10 = - 20 * Cs * (-40)

1000Cf = 800Cs

Cs = 1.25 Cf

Cs > Cf

The specific heat of the solid is more than that of the fluid.

Answer is Option (c)

Answer 2

(2)

The Latent heat of water = 540 cal/g

Mass of the water in the water in puddle = 150 g.

Amount of water Evaporated = 0.50 g

We know that

Heat lost Qw = mass of evaporated water * Latent Heat

Qw = -0.50 * 540

Qw = -270 cal.

Negative sign indicated the lost heat.

Therefore the Heat lost

Qw = mC ΔT

Where m is the mass of the Water remaining.

Remaining mass = 150- 0.5 = 149.5 g

Remaining mass = 1.0 kg.

C is the Specific Heat is 1 cal/g

ΔT is the Change in Temperature.

ΔT = (Qw / mC)

ΔT = (-270) / (149.5 * 1)

ΔT = -1.8°C

Therefore the Change in temperature is -1.8°C.

Option (B) is the correct answer.

Answer 3

(2)

The Latent heat of water = 540 cal/g

Mass of the water in the water in puddle = 150 g.

Amount of water Evaporated = 0.50 g

We know that

Heat lost Qw = mass of evaporated water * Latent Heat

Qw = -0.50 * 540

Qw = -270 cal.

Negative sign indicated the lost heat.

Therefore the Heat lost

Qw = mC ΔT

Where m is the mass of the Water remaining.

Remaining mass = 150- 0.5 = 149.5 g

Remaining mass = 1.0 kg.

C is the Specific Heat is 1 cal/g

ΔT is the Change in Temperature.

ΔT = (Qw / mC)

ΔT = (-270) / (149.5 * 1)

ΔT = -1.8°C

Therefore the Change in temperature is -1.8°C.

Option (B) is the correct answer.