An object with a mass of 3 kg, temperature of 270 ^oC, and a specific heat of 18 J/(kg*K) is dropped into a container with 36 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Feb 20, 2017

The water does not evaporate. The increase in temperature of the water =0.1ºC

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

Let $T =$ final temperature of the object and the water

For the cold water, $\Delta {T}_{w} = T - 0 = T$

For the object $\Delta {T}_{o} = 270 - T$

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(\Delta {T}_{w}\right)$

${C}_{w} = 4186 J k {g}^{-} 1 {K}^{-} 1$

${C}_{o} = 18 J k {g}^{-} 1 {K}^{-} 1$

${m}_{0} {C}_{o} \cdot \left(270 - T\right) = {m}_{w} \cdot 4186 \cdot T$

$3 \cdot 18 \cdot \left(270 - T\right) = 36 \cdot 4186 \cdot T$

$270 - T = \frac{36 \cdot 4186}{3 \cdot 18} \cdot T$

$270 - T = 2790.7 T$

$2791.7 T = 270$

T=270/2791.7=0.1ºC