Question

1)A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.1 m × 1.8 m. On a certain day, the indoor temperature is 23°C and the outdoor temperature is 0°C.

(a) What is the rate at which energy is transferred by heat
through the glass?

_________W

(b) How much energy is lost through the window in one day, assuming
the temperatures inside and outside remain constant?

_________J

2) A(n) 6.0-kg copper block is given an initial speed of 3.1 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest.

(a) If the block absorbs 85% of its initial kinetic energy as
internal energy, calculate its increase in temperature.

_____________ °C

(b) What happens to the remaining energy?

A)It becomes chemical energy.It vanishes from the universe.

B)It is so minute that it doesn't factor into the equation.

C)It is absorbed by the horizontal surface on which the block slides.

Answer #1

1)

a)

The rate of heat loss through a window glass is given by the expression,

P = -kA*dT/dx

where P is the power loss,

k is the thermal conductivity of the glass,

A is the area of the window,

dT/dx is the temperature gradient across the window.

P = -kA*(Tout - Tin) / t

where Tout is the external ambient temperature,

Tin is the internal room temperature, and t is the thickness of the glass material.

Given:

k = 0.8 W/m.°C,

A = 1.1 m × 1.80 m,

Tout = 0 °C,

Tin = 23 °C,

t = 0.62 cm = 0.62 x 10^-2 m

P = -0.8 x 1.1 x 1.8 x (0 - 23) / 0.62 x 10^-2

P = 5876.13 W ans(a)

Energy = power x time

t = 1 day = 86400 sec

E = 5876.13 W x 86400 sec

E = 5076975.5 J

E = 5.08 x 10^6 J ...........ans(b)

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