When resting, a person has a metabolic rate of about 4.83×105 joules per hour. The person is submerged neck-deep into a tub containing 1148 kg of water at 26.48oC. If the heat from the person goes only into the water, by how much will the water temperature increase after 47 minutes of immersion? (enter only the numerical portion of the temperature increase in Co)
Well, let's dive into this calculation, shall we?
To find the temperature increase, we need to use the heat transfer equation:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred by the person in 47 minutes:
47 minutes = 47/60 hours
So, the heat transferred (Q) = (4.83×10^5 J/hour) * (47/60 hour)
Now, let's find the specific heat capacity of water. The specific heat capacity of water is about 4186 J/kg·°C.
Next, we substitute the values into the equation:
Q = mcΔT
(4.83×10^5 J/hour) * (47/60 hour) = (1148 kg) * (4186 J/kg·°C) * ΔT
Now, we can solve for ΔT by rearranging the equation:
ΔT = (4.83×10^5 J/hour) * (47/60 hour) / ((1148 kg) * (4186 J/kg·°C))
Calculating this, we find that the temperature increase is approximately 0.0235 °C.
So, after 47 minutes of immersion, the water temperature will increase by about 0.0235 °C. Just a tad bit warmer, don't you think?
To find the increase in water temperature, we need to use the equation:
Q = mcΔT
Where:
Q is the heat transferred to the water,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
First, let's calculate the heat transferred to the water using the person's metabolic rate:
Metabolic rate = 4.83 × 10^5 J/h
To convert it to joules per minute, we divide it by 60:
Metabolic rate = 4.83 × 10^5 J/h ÷ 60 = 8050 J/min
Since the person is immersed for 47 minutes, the heat transferred to the water is:
Heat transferred to the water = Metabolic rate × time
= 8050 J/min × 47 min
= 378,350 J
Next, we can calculate the change in temperature using the equation Q = mcΔT:
ΔT = Q ÷ (mc)
The specific heat capacity of water, c, is approximately 4.18 J/g·°C, and the mass of the water, m, is given as 1148 kg (which we convert to grams).
Converting mass to grams:
Mass of water = 1148 kg × 1000 g/kg
= 1,148,000 g
Now, we substitute the values into the formula:
ΔT = 378,350 J ÷ (1,148,000 g × 4.18 J/g·°C)
Simplifying the equation, we find:
ΔT = 0.082 °C
Therefore, the water temperature will increase by approximately 0.082 degrees Celsius after 47 minutes of immersion.
To calculate the temperature increase of the water, we can use the formula:
Q = mcΔT
Where:
Q = heat transferred (in joules)
m = mass of the water (in kilograms)
c = specific heat capacity of water (4.184 J/g°C or 4184 J/kg°C)
ΔT = change in temperature (in °C)
First, let's convert the mass of water from grams to kilograms:
1148 kg = 1148000 grams
Now let's calculate the heat transferred from the person to the water:
Q = (4.83×10^5 J/hour) × (47 minutes / 60 minutes/hour)
= (4.83×10^5 J/hour) × (0.7833 hour)
= 3.78×10^5 J
Now let's substitute the values into the equation and solve for ΔT:
3.78×10^5 J = (1148000 kg) × (4184 J/kg°C) × ΔT
Simplifying the equation:
3.78×10^5 J = 4.8×10^9 J/°C × ΔT
Dividing both sides by 4.8×10^9 J/°C:
ΔT = 3.78×10^5 J / (4.8×10^9 J/°C)
= 0.07875 °C
Therefore, the water temperature will increase by approximately 0.07875 °C after 47 minutes of immersion.