Question

100.0 g of water was placed in a simple, constant-pressure calorimeter. The temperature of the water was recorded as 295.0 K. A 20.0 g copper block was heated to 353.0 K and then dropped into the water in the calorimeter. What was the final temperature of the water if the specific heat capacities of copper is 0.385 J/g K

Answer 1

Answer:

$296.05\ \text{K}$

Explanation:

$m_w$ = Mass of water = 100 g

$c_w$ = Specific heat of water = $4.184\ \text{J/g K}$

$m_c$ = Mass of copper = 20 g

$c_c$ = Specific heat of copper = $0.385\ \text{J/g K}$

$\Delta T_w$ = Temperature change in water = $(T-295)$

$\Delta T_c$ = Temperature change in cooper = $(353-T)$

T = Final temperature of the system

The heat balance of the system is given by

$m_wc_w\Delta T_w=m_cc_c\Delta T_c\\\Rightarrow 100\times 4.184\times (T-295)=20\times 0.385\times (353-T)\\\Rightarrow 418400\left(T-295\right)=7700\left(353-T\right)\\\Rightarrow 418400T-123428000=2718100-7700T\\\Rightarrow T=\frac{1261461}{4261}\\\Rightarrow T=296.05\ \text{K}$

The final temperature of the water is $296.05\ \text{K}$.