Question

A 100 g glass container is at 10.0°C. 200 g of water at 90.0°C is added to the glass container. What is the final temperature of the water and the glass, in °C? (specific heat of water = 1.00 cal/g °C, specific heat of glass = .200 cal/gm °C)

Answer 1

mass of glass container, m = 100g

temperature of glass container = 10°

mass of water , M = 200g

temperature of water = 90°

we know, heat flows through higher temperature to lower temperature.

it means after adding water into glass container, heat lost by water = heat gained by glass container.

Let final temperature is T.

heat lost by water = MS(90 - T)

here M = 200g, S = 1cal/g.°C

then, heat lost by water = 200 × 1(90- T)

= 200(90 - T)......(1)

heat gained by glass container = ms(T - 10)

here, m = 100g, s = 0.2cal/g.°C

so, heat lost by glass container = 100 × 0.2(T - 10)

= 20(T - 10).....(2)

from equations (1) and (2),

200(90 - T) = 20(T - 10)

or, 10(90 - T) = T - 10

or, 900 - 10T = T - 10

or, 910 = 11T

or, T = 910/11 = 82.72°C