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 $296.05\ \text{K}$.

Answer 2

The final temperature of the water when placed in a calorimeter is 296.05K

HOW TO CALCULATE FINAL TEMPERATURE:

• The final temperature of water placed in a calorimeter can be calculated using the following expression:

• Q(water) = - Q(copper)

• (m × c × ∆T) water = - {m × c × ∆T} copper

Where;

1. Mass of water = 100 g
2. Specific heat of water = 4.184 J/g K
3. Mass of copper = 20 g
4. Specific heat of copper = 0.385 J/g K
5. Temperature change in water = T - 295K
6. Temperature change in copper = T - 353K

• 100 × 4.184 × (T - 295) = - {20 × 0.385 × (T - 353)}

• 418.4T - 123428 = - (7.7T - 2718.1)

• 418.4T - 123428 = -7.7T + 2718.1

• 418.4T + 7.7T = 123428 + 2718.1

• 426.1T = 126146.1

• T = 126146.1 ÷ 426.1

• T = 296.05K

• Therefore, the final temperature of the water when placed in a calorimeter is 296.05K.

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