Blood temperature drop from heat loss and heating water calculations

[moonlit]

I'm stuck on a few problems that I was assigned for homework. Not sure if someone can help me with these...

1) Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, 0.645 kg of blood flows to the surface of the body and releases 1860 J of energy. The blood arriving at the surface has the temperature of the body interior, 37.3 °C. Assuming that blood has the same specific heat capacity as water, determine the temperature in degrees Celsius of the blood that leaves the surface and returns to the interior.

2) If the price of electrical energy is 0.174 dollars per kilo-watt hour, what is the dollar cost of using electrical energy to heat the water in a swimming pool (13.3 m x 11.3 m x 1.93 m) from 11.3 to 26.9 °C?

I know that the amount of heat needed to heat the water is 2.3x10^7 J and that one kilowatt per hour = 3.60x10^6 J. Which means that the number of kilowatthours of energy used to heat the water is 6.4 kwH. Just not sure what equation to use for the size of the swimming pool and the temp difference

3) When resting, a person has a metabolic rate of about 2.34 x 105 joules per hour. The person is submerged neck-deep into a tub containing 1.81 x 103 kg of water at 26.8 °C. If the heat from the person goes only into the water, find the water temperature in degrees Celsius after half an hour.

 

[moonlit]

Ahhhh forget number 2...I think I know where I went wrong there. Copied down the wrong numbers.

 

[M98Ranger]

For the first problem, we can use the formula Q = mcΔT, where Q is the energy released by the blood, m is the mass of blood, c is the specific heat capacity of water, and ΔT is the change in temperature. We can rearrange the formula to solve for the final temperature, which gives us ΔT = Q/(mc). Plugging in the given values, we get ΔT = 1860 J / (0.645 kg x 4186 J/kg·°C) = 0.007 °C. Since the initial temperature was 37.3 °C, the final temperature of the blood returning to the interior is 37.3 °C - 0.007 °C = 37.293 °C.

For the second problem, we can use the formula W = Pt, where W is the energy used, P is the power in watts, and t is the time in hours. We can convert the volume of the pool from cubic meters to liters (13.3 m x 11.3 m x 1.93 m = 285.9 m^3 = 285,900 L) and then use the formula V = mρ, where V is the volume, m is the mass, and ρ is the density of water. This gives us the mass of water in the pool as 285,900 L x 1 kg/L = 285,900 kg. Then, we can use the formula Q = mcΔT to calculate the energy needed to heat the water from 11.3 °C to 26.9 °C, which gives us Q = (285,900 kg) x (4186 J/kg·°C) x (26.9 °C - 11.3 °C) = 2.3 x 10^7 J. Finally, we can plug in the given values into the formula W = Pt, which gives us W = (6.4 kW) x (t) = 2.3 x 10^7 J. Solving for t, we get t = (2.3 x 10^7 J) / (6.4 kW) = 3593.75 seconds = 59.9 minutes. Therefore, the cost of using electrical energy to heat the water in the swimming pool for 59.9 minutes is 0.174 dollars/kWh x