Question

An object of mass$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{Kg}$falls through a height ofand $\mathbf{50}\mathbf{.0}\mathbf{}\mathbf{\text{m}}$, by means of a mechanical linkage, rotates a paddle wheel that stirs$\mathbf{0}\mathbf{.600}\mathbf{}\mathbf{\text{kg}}$of water. Assume that the initial gravitational potential energy of the object is fully transferred to thermal energy of the water, which is initially at$\mathbf{15}\mathbf{.0}\mathbf{\text{°C}}$.What is the temperature rise of the water?

Short answer

The rise in the temperature of water is$1.17\text{°C}$

Step-by-step solution

Stating thegiven data

1. Mass of an object is,$m=6\text{\hspace{0.17em}kg}$
2. Mass of water is,$m_w=0.600\text{kg}$
3. The initial height of an object is,$h=50.0\text{\hspace{0.17em}m}$
4. The initial temperature of water,$T_i=15\text{°C}$.

Understanding the concept of energy

We need to equate the equations of gravitational potential energy and thermal energy, as the potential energy is converted to thermal energy.

Formulae:

Potential energy of the body$PE=mgh$,…(i)

Heat transferred by the body via thermal radiation$Q=mc\Delta t$,…(ii)

where c is specific heat capacity of water = .$4182\text{\hspace{0.17em}J/kg\hspace{0.17em}°C}$

Calculation of the rise in temperature

From the equation (i) and the given values, we get the potential energy as

$\begin{array}{c}PE=\left(6.00\text{\hspace{0.17em}kg}\right)\times \left(9.81\text{\hspace{0.17em}\hspace{0.17em}N/kg}\right)\times \left(50.0\text{\hspace{0.17em}m}\right)\\ =2943\text{J}\end{array}$

We are given that the gravitational potential energy is fully transferred to thermal energy of water. So, PE = Q. Thus, using the value of above potential energy and equation (ii), the rise in temperature is given by

$\begin{array}{c}2943\text{\hspace{0.17em}J}=m_w\times c\times \Delta t\\ 2943\text{\hspace{0.17em}J}=\left(0.600\text{\hspace{0.17em}kg}\right)\times \left(4182\text{\hspace{0.17em}J/kg\hspace{0.17em}°C}\right)\times \Delta t\text{(substitutingthegivenvalues)}\\ \Delta t=\frac{\left(2943\text{\hspace{0.17em}J}\right)}{\left(0.600\text{\hspace{0.17em}kg}\right)\times \left(4182\text{\hspace{0.17em}J/kg\hspace{0.17em}°C}\right)}\\ =1.17\text{°C}\end{array}$

Hence, the value of the rise in temperature is.$1.17\text{°C}$