# A 32 gram sample of vanadium was heated to 75°C, and placed in a calorimeter containing water at a temperature of 22.5°C. The final temperature of both the metal and the water was 26.3°C. What is the mass of the water in the calorimeter?

Mar 4, 2016

The mass of the water is 48 g.

#### Explanation:

This is a calorimetry problem.

There are two heat flows involved.

$\text{heat of vanadium + heat of water = 0}$

${q}_{1} + {q}_{2} = 0$

m_1c_1ΔT + m_2c_2ΔT = 0

${m}_{1} = \text{32 g"; c_1 = "0.489 J·°C"^"-1""g"^"-1"; ΔT = "(26.3- 75) °C" = "-48.7 °C}$
m_2 = ?;color(white)(ml)c_2 = "4.18 J·°C"^"-1""g"^"-1"; color(white)(l)ΔT = "(26.3 - 22.5) °C" = "3.8 °C"

q_1 = m_1c_1ΔT = 32 color(red)(cancel(color(black)("g"))) × "0.489 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × ("-48.7") color(red)(cancel(color(black)("°C"))) = "-762 J"

q_2 = m_2c_2ΔT = m_2 × 4.18 "J·"color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 3.8 color(red)(cancel(color(black)("°C"))) = 15.9color(white)(l) m_2color(white)(l) "J·g"^"-1"

${q}_{1} + {q}_{2} = \text{-762" color(red)(cancel(color(black)("J"))) + 15.9 color(white)(l)color(white)color(red)(cancel(color(black)("J")))"·g"^"-1} = 0$

${\text{-762" + 15.9 m_2color(white)(l) "g}}^{-} 1 = 0$

m_2 = 762/("15.9 g"^"-1") = "48 g"