Question

A 90 kg person jumps from a 30 m tower into a tub of water with a volume of 5 m^3 initially at 20 degrees C. Assuming that all of the work done by the person is converted into heat to the water, what is the final temperature of the water?

Answer 1

Let:

$$P = 90kg$$ $$h = 30m$$ $$t1 = 20{}^\circ \text{C}$$ $$C(\text{water}) = 1,82 \frac{KJ}{KG{}^\circ\text{C}}$$ $$V(\text{water}) = 5m^3$$ $$M(\text{water}) = \rho Vg = 1000 \times 5 \times 9.8 = 49000N$$ $$t2 = ?$$

The work done by person will be as: $A = mgh = Ph$

$$A = MC(\Delta t)$$ $$\Delta t = \frac{A}{MC} = \frac{Ph}{MC}$$ $$t2 = t1 + \Delta t$$ $$t2 = t1 + \frac{PH}{MC}$$ $$t2 = 20 + \frac{90 \times 30}{49000 \times 1.82} = 20.06{}^\circ \text{C}$$

Answer: $t2 = 20.06{}^\circ \text{C}$