Question

A block of tin with a mass of 1.70 kg, initially at a
temperature of 150.0°C, is in a well-insulated container. Water at
a temperature of 26.0°C is added to the container, and the entire
interior of the container is allowed to come to thermal
equilibrium, where it reaches a final temperature of 61.0°C. What
mass of water (in kg) was added? Assume any water turned to steam
subsequently recondenses.

Answer #1

Since heat generate by temperature change is given by,

Q = m*C*dT

Given that final temperature is 61.0 degC

Now using energy conservation:

Heat released by metal = Heat gained by liquid

Q1 = Q2

Mm*Cm*dT1 = M*C*dT2

Mm = mass of metal = 1.70 kg

M = mass of water = ??

Cm = specific heat capacity of tin = 210 J/kg-C

C = specific heat capacity of water = 4186 J/kg-C

dT1 = 150 - 61.0 = 89.0

dT2 = 61.0 - 26.0 = 35.0

Using these values:

1.70*210*89.0 = M*4186*35.0

M = (1.70*210*89.0)/(4186*35.0)

**M = 0.217 kg**

"Let me know if you have any query."

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