91Ó°ÊÓ

1. Mass of an object is,$m=6\text{ k²µ}$
2. Mass of water is,$m_w=0.600\text{kg}$
3. The initial height of an object is,$h=50.0\text{″¾}$
4. The initial temperature of water,$T_i=15\text{°°ä}$.

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$PE=mgh$,…(¾±)

Heat transferred by the body via thermal radiation$Q=mc\mathrm{Δ}t$,…(¾±¾±)

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

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

$\begin{array}{c}PE=\left(6.00\text{ k²µ}\right)×\left(9.81\text{ â¶Ä‰N/°ì²µ}\right)×\left(50.0\text{″¾}\right)\\ =2943\text{J}\end{array}$

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

$\begin{array}{c}2943\text{ J}=m_w×c×\mathrm{Δ}t\\ 2943\text{ J}=\left(0.600\text{ k²µ}\right)×\left(4182\text{ J/kg °°ä}\right)\text{ }×\mathrm{Δ}t\text{(substitutingthegivenvalues)}\\ \mathrm{Δ}t=\frac{\left(2943\text{ J}\right)}{\left(0.600\text{ k²µ}\right)×\left(4182\text{ J/kg °°ä}\right)}\\ =1.17\text{°°ä}\end{array}$

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