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Zanzabum
3 years ago
8

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
Chemistry
1 answer:
Marizza181 [45]3 years ago
7 0

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}.

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