dariushjhp0ckhk
contestada

A 1.00x104 g piece of metal at 50.0°C is placed in 1000. g of water at 10.0°C. The metal and water come to the same temperature of 31.4 °C. What is the specific heat of the metal? Find the specific heat of the pure metal in both cal/(g·°C) and J/(g·°C).

Respuesta :

prosborough198p2ub41
c=Q/m*deltaT
Qwater=1000g*4.184J/gC*21.4C=89537.6J
cmetal=89537.6J/(10000g)*(18.6C)=0.481J/g*C  
Qwater=1000g*1.00cal/C*21.4=21400cal
cmetal= 21400cal/(10000g)*(18.6C)=0.115cal/g*C

Hope this helps you understand the concept! Ant questions please feel free to ask! Please rate brainliest if my answer helped you! Thank you so much!!

Otras preguntas