Question

A 312 g sample of a metal is heated to 277.845 °C and plunged into 200 g of water at a temperature of 11.945 °C. The final temperature of the water is 99.062 °C. Assuming water has a specific heat capacity of 4.184 J/g °C, what is the specific heat capacity of the metal sample, in J/g °C)? Assume no heat loss to the surroundings.

Answer 1

Answer:

The specific heat capacity of the metal is 1.307 J/g °C

Explanation:

Step 1: Data given

mass of the metal = 312 grams

initial temperature of the metal ( before plunged in the water) = 277.845 °C

initial temperature water = 11.945 °C

Final temperature of water (and aslo the metal) = 99.062 °C

Specific heat capacity of wayer = 4.184 J/g °C

Step 2:  Calculate the specifi heat capacity of the metal

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

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

with mass of metal = 312 grams

with C(metal) = TO BE DETERMINED

with ΔT (metal) = 99.062 - 277.845 = -178.783 °C

with mass of water = 200 grams

with C(water) = 4.184 J/°C * g

with ΔT (water) = 99.062 - 11.945 = 87.117 °C

312 * C(metal) * -178.783  = - 200* 4.184 * 87.117

C(metal) = -72899.51 / 55780.296 = 1.307

The specific heat capacity of the metal is 1.307 J/g °C