Question

hot water at 100°c is added to 300g of water at 0°c until the final temperature is 40 degree Celsius.find the mass of a hot water added.(the specific heat capacity of water is 4200J/kg°c)​

Answer 1

Answer:

$m_h=0.2$

Explanation:

Using the heat formula and concept of calorimetry to answer this problem.

$\boxed{\left\begin{array}{ccc}\text{\underline{Formula for Heat:}}\\\\Q=mc\Delta T\end{array}\right}$

$\boxed{\left\begin{array}{ccc}\text{\underline{Calorimetry (Energy Conservation):}}\\Q_{gain}+Q_{lose}=0\end{array}\right}$

Given:

$T_{0_{h}}=100 \textdegree C\\T_{0_{c}}=0 \textdegree C\\T_{f}=40 \textdegree C\\m_c=300 \ g \rightarrow 0.3 \ kg\\c_w=4200 \ \frac{J}{kg\textdegree C}$

Find:

$m_h= \ ?? \ kg$

(1) - Identify what is gaining heat and what is losing heat and form an equation

The cold water is gaining heat from the added hot water, which means the hot water is losing heat.

$Q_{gain}=m_cc_w(T_f-T_{0_{c}})\\\\Q_{lose}=m_hc_w(T_f-T_{0_{h}})\\\\\therefore \boxed{m_cc_w(T_f-T_{0_{c}})+m_hc_w(T_f-T_{0_{h}})=0}$

(2) - Plug the known values into the equation and solve for the unknown "m_h"

$m_cc_w(T_f-T_{0_{c}})+m_hc_w(T_f-T_{0_{h}})=0\\\\\Longrightarrow (0.3)(4200)(40-0)+m_h(4200)(40-100)=0\\\\\Longrightarrow (0.3)(40)-m_h(60)=0\\\\\Longrightarrow m_h(60)=(0.3)(40)\\\\\therefore \boxed{\boxed{m_h=0.2 \ kg}}$

Thus, the mass of the hot water is found.