Answer to Question #123524 in General Chemistry for Ingrid

Question

sample of metal with a mass of 54.3 g at 100.0 ºC is dropped into 175.0 g of water at 25.3 ºC. The temperature of the water rises to 27.1 ºC. Assume there is no heat lost to the environment. Note: The specific heat of water is 4.184 J/gºC.
What is the amount of heat gained by the water?

ii) What is the specific heat of the metal?

Answer 1

According to the heat equation:

Q = m × c × (T₂ - T₁)

where Q - gained or lost heat, m - mass, c - specific heat, T₂ - final temperature, T₁ - initial temperature.

From here, the amount of heat gained by the water equals:

Q_water = m_water × c_water × (T₂ - T₁) = 175.0 g × 4.184 J/gºC × (27.1 ºC - 25.3 ºC) = 1317.96 J = 1.32 kJ

As there is no heat loss in the system, the same heat was lost by the metal dropped into the water while its final temperature equals 27.1 ºC. As a result:

Q_metal = m_metal × c_metal × (T₂ - T₁)

From here, the specific heat of the metal is:

c_metal = Q_metal / [m_metal × (T₂ - T₁)] = -1317.96 J / [54.3 g × (27.1 ºC - 100.0 ºC)] = 0.333 J / gºC

Answer: 1.32 kJ and 0.333 J / gºC