In: Physics
A cold glass of ginger ale warms 99% of the way from 0 °C to 20 °C in 50 minutes. This means the initial warming is at a rate of 2 °C per minute 5,
A) If the glass is modeled as 350 ml of liquid water, what is the initial rate of heat flow into the glass?
B) Frozen grapes can be used to keep a beverage cold without diluting it. If 30 g of grapes are added, and they start at -20 °C, how much heat wil it take to turn them into liquid water at 0 °C?
C)Assuming the same rate of heat flow as found in (a), how much time will (b) take? This is how long the grapes can keep the drink cold a. b. c.
D) Reverse the logic of the question. If adding 30 g of ice can hold the drink at 0 C for 60 minutes (at which point, the ice is melted), find the initial warming rate (degrees per minute) once the ice is finished melting. (Assume constant heat flow from the environment during the melting of the ice.)
specific heat capacity of water = 4.186 J/g; take density of water to be approxiately 1 g/ml. Thus weight of 350 mL of water = 350 g.
(a)The heat transfer increses temperature of water. The heat transfer rate i.e. rate of heat transfer can be taken from rate of temperature change.
as rate of temperature rise is 2 degrees in 5 minutes (5 minutes = 5*60 = 300 s)
or 9.77 W
(b) The assumption is that the heat capacity of grapes is neglected.
First the ice at -20 degree C will turn to ice at 0 degree C and then water at 0 degree C
specifice heat of ice = 2.09 J/g/degree C
latent heat of ice = 334 J/g
thus heat required =
(c) for rate of heat supply at 9.77 J/s, the time in seconds to supply 131530 J of energy is given by the equation
(d) for latent heat of fusion = 334 J/g, the amount of energy for 30 g of ice to melt
thus rate of heat flow to melt ice, i.e to supply 10020 J of energy in 60 min or 3600 s = 10020/3600 = 2.78 J/s
Now after ice has melted we have 380 g of water at 0 degree C. and heat is supplied at rate of 2.78 J/s. We have find rate of change of temperature.
degrees per second degrees per minute (ans)