Question

A 30.5 g sample of an alloy at 95.0°C is placed into 49.3 g water at 24.3°C in an insulated coffee cup. The heat capacity of the coffee cup (without the water) is 9.2 J/K. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy? (c of water is 4.184 J/g×K)

Answer 1

0.752 J/g*K

Step-by-step explanation:

The heat lost by the alloy (which is negative) must be equal to the heat gained by the water and the coffee cup:

-Qa = Qw + Qc

-ma*ca*ΔTa = mw*cw*ΔTw + C*ΔTc

Where, m is the mass, c is the specific heat capacity, C is the heat capacity of the coffee cup, ΔT is the change in temperature, a represents the alloy, and w the water.

The coffee cup has initial temperature equal to the water, then:

-30.5*ca*(31.1 - 95.0) = 49.3*4.184*(31.1 - 24.3) + 9.2*(31.1 - 24.3)

1948.95ca = 1465.20

ca = 0.752 J/g*K