A 50-lb iron ball is heated to 200°F and is then immediately plunged into a vessel containing 100 lb of water whose temperature is 40°F. The specific heat of iron is 0.11. (a) Find the temperature of the body as a function of time. Hint. The quantity of heat lost by the iron body in time t is 50(0.11) (200 - TB), where TB is its temperature at the end of time t. The quantity of heat gained by the water-remember the specific heat of water is one- is 100(1)(Tw 40), where Tw is the temperature of the water at the end of time t. Since the heat gained by the water is equal to the heat lost by the ball, 50(0.11) (200- 100(Tw40). Solve for Tw and substitute this value for TM in (15.3). Solve for TB. (b) Find the common temperature approached by body and water as t→→ ∞. TB) - -
A 50-lb iron ball is heated to 200°F and is then immediately plunged into a vessel containing 100 lb of water whose temperature is 40°F. The specific heat of iron is 0.11. (a) Find the temperature of the body as a function of time. Hint. The quantity of heat lost by the iron body in time t is 50(0.11) (200 - TB), where TB is its temperature at the end of time t. The quantity of heat gained by the water-remember the specific heat of water is one- is 100(1)(Tw 40), where Tw is the temperature of the water at the end of time t. Since the heat gained by the water is equal to the heat lost by the ball, 50(0.11) (200- 100(Tw40). Solve for Tw and substitute this value for TM in (15.3). Solve for TB. (b) Find the common temperature approached by body and water as t→→ ∞. TB) - -
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Solve the differential equation, thank you.
Transcribed Image Text:The specific heat of a substance is defined as the ratio of the quantity
of heat required to raise a unit weight of the substance 1º to the quantity
of heat required to raise the same unit weight of water 1°. For example, it
takes 1 calorie to change the temperature of 1 gram of water 1°C (or 1
British thermal unit to change the temperature of 1 lb of water 1°F). If,
therefore, it takes only of a calorie to change the temperature of 1 gram
of a substance 1°C (or of a British thermal unit to change 1 lb of the
substance 1°F), then the specific heat of the substance is .
In problems 5-7 below, assume that the only exchanges of heat occur
between the body and water.
5. A 50-lb iron ball is heated to 200°F and is then immediately plunged into a
vessel containing 100 lb of water whose temperature is 40°F. The specific
heat of iron is 0.11. (a) Find the temperature of the body as a function of
time. Hint. The quantity of heat lost by the iron body in time t is 50(0.11)
(200 TB), where TB is its temperature at the end of time t. The quantity
of heat gained by the water-remember the specific heat of water is one-
is 100(1)(Tw 40), where Tw is the temperature of the water at the end
of time t. Since the heat gained by the water is equal to the heat lost by the
ball, 50(0.11) (200 TB) 100(Tw 40). Solve for Tw and substitute
this value for TM in (15.3). Solve for TB. (b) Find the common temperature
approached by body and water as t → ∞.
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