Question

What is the specific heat of a metal if a ten gram sample at 40.2°C is dropped into 100 grams of water at 25.4°C and the temperature rises to 28.6°C? (Sp. heat of H_2O = 4.18 J/g•°C)

a. 2.11 J/g °C
b. 0.085 J/g °C
c. 206 J/g °C
d. 1.92 J/g °C
e. 20.9 J/g °C

Answer 1

The specific heat of the metal is 2.11 J/g°C.

Step-by-step explanation:

The specific heat of a metal can be calculated using the equation q = mc∆T, where q is the heat transferred, m is the mass, c is the specific heat, and ∆T is the change in temperature. In this case, the heat transferred to the water can be calculated as follows:

• Heat transferred to water = (mass of water) x (specific heat of water) x (change in temperature)
• Heat transferred to water = (100 g) x (4.18 J/g•℃) x (28.6 ℃ - 25.4 ℃)
• Heat transferred to water = 1209.6 J

The heat transferred to the metal is equal to the heat transferred to the water, so we can use the same equation to calculate the specific heat of the metal:

• (specific heat of metal) = (heat transferred to metal) / (mass of metal x change in temperature)
• (specific heat of metal) = 1209.6 J / (10 g x (40.2 ℃ - 28.6 ℃))
• (specific heat of metal) = 2.11 J/g•℃

Therefore, the specific heat of the metal is 2.11 J/g•℃.