A student needs to determine the specific heat of a metal. She sets up a calorimeter with 100.0 mL of deionized water at room temperature, 21.0 Celsius. She heats up 50.0g of the unknown metal to 100.0 Celsius. After adding the hot metal to the water in the calorimeter the temperature equilibrated at 28.7 Celsius. What is the specific heat of the metal in J/gC?

The specific heat of the metal is 0.903 J/g°C

Explanation:

Step 1: Data given

Volume water = 100.0 mL = 0.100 L

Temperature = 21.0 °C

Mass of metal = 50.0 grams

Initial temperature of metal = 100.0 °C

Final temperature = 28.7 °C

The specific heat capacity of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Heat lost = heat gained

Qlost = -Qgained

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal)*ΔT(metal) = -m(water) * c(water) * ΔT(water)

⇒with m(metal) = the mass of metal = 50.0 grams

⇒with c(metal) = the specific heat of the metal = TO BE DETERMINED

⇒with ΔT(metal) = The change of temperature of the metal = T2 - T1 = 28.7 °C - 100 °C = -71.3 °C

⇒with m(water) = the mass of water = 100.0 ml * 1g/mL = 100 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the chagnge of temperature of water = 28.7 °C - 21.0 °C = 7.7 °C

50.0 * c(metal) * (-71.3) = -100 * 4.184 * 7.7 °C

-3565 * c(metal) = -3221.7

c(metal) = 0.903 J/g°C

The specific heat of the metal is 0.903 J/g°C

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .