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natali 33 [55]
3 years ago
11

An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

ture of the water in the cup was 19.5c. what was the mass of ice cube?
Physics
1 answer:
Ede4ka [16]3 years ago
3 0
The amount of heat given by the water to the block of ice can be calculated by using
Q=m_w C_{sw} \Delta T_w
where 
m_w = 30 g is the mass of the water
C_{sw}=4.18 J/(g ^{\circ}C) is the specific heat capacity of water
\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C is the variation of temperature of the water.

Using these numbers, we find
Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
Q = m_i L_f
where m_i is the mass of the ice while L_f =334 J/g is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
m_i =  \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g
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