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worty [1.4K]
3 years ago
11

If a 95.0 gram sample of metal at 100.0 oC is placed in 50.0 g of water with an initial temperature of 22.5oC and the final temp

erature of the system is 48.5oC, what is the specific heat of the metal? Group of answer choices
Chemistry
1 answer:
kaheart [24]3 years ago
8 0

Answer : The specific heat of the metal is, 1.11J/g^oC

Explanation

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of metal = 95.0g

m_2 = mass of water  = 50.0 g

T_f = final temperature of mixture = 48.5^oC

T_1 = initial temperature of metal = 100.0.0^oC

T_2 = initial temperature of water = 22.5^oC

Now put all the given values in the above formula, we get

(95.0g)\times c_1\times (48.5-100.0)^oC=-[(50.0g)\times 4.18J/g^oC\times (48.5-22.5)^oC]

c_1=1.11J/g^oC

Therefore, the specific heat of the metal is, 1.11J/g^oC

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