A block of tin with a mass of 1.90 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 64.0°C. What mass of water (in kg) was added? Assume any water turned to steam subsequently recondenses.

When the temperature of an object of specific heat capacity [tex]c[/tex] and mass [tex]m[/tex] changes by [tex]\Delta T[/tex], the associated energy change would be:

[tex]Q = c\, m\, \Delta T[/tex].

Rearranging this equation would give an expression for the mass of the object given the energy change, specific heat, and temperature change:

[tex]\displaystyle m = \frac{Q}{(c)\, (\Delta T)}[/tex].

In this question:

Temperature change of the block of tin ([tex]m= 1.90\; {\rm kg}[/tex]): [tex]\Delta T = (150 - 64.0)\; {\rm K}[/tex].
Temperature change of the water added to the container: [tex]\Delta T = (64.0 - 26.0)\; {\rm K}[/tex]

Apply the following steps to find the mass of water added:

Find the energy that the block of tin released using the equation [tex]Q = c\, m\, \Delta T[/tex].
Since energy conserved and the container is insulated, the energy water in the container absorbed would be equal to the energy the tin block released.
Apply the following equation the find the mass of the water added:
[tex]\displaystyle m = \frac{Q}{(c)\, (\Delta T)}[/tex].

Look up the specific heat capacity of tin and water:

Tin: approximately [tex]213\; {\rm J\cdot kg^{-1} \cdot K^{-1}}[/tex].
Water: approximately [tex]4\, 186\; {\rm J\cdot kg \cdot K^{-1}}[/tex].

The energy that the block of tin released would be:

[tex]\begin{aligned} Q &= c\, m\, \Delta T \\ &= (213\; {\rm J\cdot kg^{-1} \cdot K^{-1}})\, (1.90\; {\rm kg})\, ((150 - 64.0)\; {\rm K}) \\ &\approx 34\, 800\; {\rm J}\end{aligned}[/tex].

Since the container is insulated, under the assumptions, energy water in the container absorbed would be equal to the energy the block of tin released: [tex]34\, 800\; {\rm J}[/tex]. Given that the temperature change of the water is [tex]\Delta T = (64.0 - 26.0)\; {\rm K}[/tex] and that the specific heat of water is approximately [tex]4\, 186\; {\rm J\cdot kg \cdot K^{-1}}[/tex], the mass of water added to the container would be:

[tex]\begin{aligned} m &= \frac{Q}{(c)\, (\Delta T)} \\ &\approx \frac{34\, 800\; {\rm J}}{(4\, 186\; {\rm J\cdot kg \cdot K^{-1}})\, ((64.0 - 26.0)\; {\rm K})} \\ &\approx 0.219\; {\rm kg}\end{aligned}[/tex].

In other words, the mass of water added to the container should be approximately [tex]0.219\; {\rm kg}[/tex].