Question

94.0 g of a metal at 88.0°C are added to 51.0 g of water at 34.4°C. When the system reaches constant temperature, the temperature is 37.5°C. What is the specific heat of the metal? The specific heat of water is 4.184 J/g·°C. Calculate ΔE for the process. Include the sign.

Answer 1

Cp = 0.139 J/g·°C

ΔE = -659.83 J

Step-by-step explanation:

This is a problem of thermo-equilibrium, so we use the formula:

• Q = m*ΔT*Cp

• Where m is mass, ΔT is the difference in temperature(Tfinal-Tinitial), and Cp is the specific heat.

Heat loss of Metal = Heat gain of water

• -Qm = Qw

We put the data given by the problem:

• - 94.0 g * (37.5-88.0)°C * Cp = 51.0 g * (37.5-34.4)°C * 4.184 J/g·°C

And we solve for Cp:

• - 94.0 g * -50.5°C * Cp = 51.0 g * 3.1°C * 4.184 J/g·°C

• 94.0 g * 50.5°C * Cp = 661.49 J

• Cp = 0.139 J/g·°C

In order to calculate ΔE we the Cp of metal, its mass and ΔT:

• ΔE = -Cp * m * ΔT

• ΔE = -659.83 J