A 65.0 gram sample of some unknown metal at 100.0° C is added to 177.1 grams of water at 24.5° C. The temperature of the water rises to 27.0° C. If the specific heat capacity of liquid water is 4.18 J/ (°C × g), what is the specific heat of the metal?

0.285 J/ (°C × g).

0.390 J/ (°C × g).

1.05 J/ (°C × g).

2.56 J/ (°C × g).

Question

A 65.0 gram sample of some unknown metal at 100.0° C is added to 177.1 grams of water at 24.5° C. The temperature of the water rises to 27.0° C. If the specific heat capacity of liquid water is 4.18 J/ (°C × g), what is the specific heat of the metal?

 0.285 J/ (°C × g).

 0.390 J/ (°C × g).

 1.05 J/ (°C × g).

 2.56 J/ (°C × g).

jfraser · Answer

plug into the equation for calorimetry:$$Q = m*C_p* \Delta T$$ solve for C and the closest one wins.

anonymous · Answer

what does Q equal? M is 65.0 and delta t is 27.0 right?

jfraser · Answer

You have a hot piece of metal cooling down in water, and you have the water heating up. The amounts of heat released by the metal must be equal to the amount of heat absorbed by the water. Solve for the Q of water first, then set that Q equal to the Q of the metal, and solve for C of the metal$$Q_{H_{2}O} = m_{H_{2}O} *C_{H_{2}O}* \Delta T _{H_{2}O}$$$$Q_{metal} = m_{metal} * C_{metal} * \Delta T _{metal}$$$$Q_{H_2O} = Q_{metal}$$