Calculate the specific heat of a metal (in calories/gram-degree C) from the following data. A container made of the metal has a mass of 3.77 kg and contains 18.1 kg of water. A 1.45 kg piece of the same metal, initially at a temperature of 164 degrees C, is placed in the water. The container and water initially have a temperature of 15 degrees C, and the final temperature of the entire system is 18 degrees C.

Step-by-step explanation:

Let, the specific heat of the container is x calories/gram-degree C.

The container and water gains (18 - 15) = 3 degrees C.

Hence, the transfer of heat is .

The metal, which is dropped in the water, losses (164 - 18)= 144 degrees C.

Hence, the transfer of heat is .

As per the given conditions,

.

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg