Question

A 0.500 kg mass of unknown metal at 100.0 oC is placed in 120.0 g H2O at 25.0 oC. The final temperature of the water and metal is 30.0 oC. Ignoring the container, what is the specific heat of the metal? ( c-water = 4.184 J/goC)

A. 0.0071 J/goC
B. 0.0142 J/goC
C. 0.071 J/goC
D. 0.142 J/goC
E. None of the Above

Answer 1

Question #78025, Chemistry / General Chemistry

A 0.500 kg mass of unknown metal at 100.0 oC is placed in 120.0 g H2O at 25.0 oC. The final temperature of the water and metal is 30.0 oC. Ignoring the container, what is the specific heat of the metal? (c-water = 4.184 J/goC)

A. 0.0071 J/goC

B. 0.0142 J/goC

C. 0.071 J/goC

D. 0.142 J/goC

E. None of the Above

Solution:

$$q = m \times c \times \Delta T \text{ -heat capacity formula;}$$

1. Find heat capacity for water:

$$q(\text{water}) = m \times c \times \Delta T = 120.0g \times 4.184J/\text{goC} \times (30.0\text{oC} - 25.0\text{oC}) = 2486.4J$$

2. $q(\text{water}) = q(\text{unknown metal})$

$$q(\text{metal}) = m \times c \times \Delta T$$ $$c = \frac{q}{m \cdot \Delta T} = \frac{2486.4J}{500g \cdot (100.0\text{oC} - 30.0\text{oC})} \approx 0.071\text{J/goC}$$

Answer: C