# An object with a mass of 64 g is dropped into 210 mL of water at 0^@C. If the object cools by 96 ^@C and the water warms by 7 ^@C, what is the specific heat of the material that the object is made of?

Jul 14, 2017

$1.00 \cdot {10}^{- 3} k J k {g}^{- 1} {K}^{- 1}$

#### Explanation:

Object:
Mass - $64 g = 0.064 k g$
Specific heat cpacity - ?
Temperature change = $96 K$

Water:
Mass - $210 m L = 0.00021 {m}^{3} \approx 0.00021 k g$.
Specific heat capacity - $4.18 k J {K}^{- 1} k {g}^{- 1}$
Temperature change = $7 K$

${m}_{o} {C}_{o} \Delta {T}_{o} = {m}_{w} {C}_{w} \Delta {T}_{w}$

$\left(0.064 \cdot 96\right) {C}_{o} = \left(4.18 \cdot 7 \cdot 0.00021\right)$

${C}_{o} = \frac{4.18 \cdot 7 \cdot 0.00021}{0.064 \cdot 96} = 0.00100009766 \approx 1.00 \cdot {10}^{- 3} k J k {g}^{- 1} {K}^{- 1}$