Question

50 g of copper is heated to increase its temperature by 10° C. If the same quantity of heat is given to 5 g water, the rise in its temperature is [Specific heat of copper = 420 joule-kg^-1 °C^-1 , specific heat of water = 4200 joule-kg^-I °C^-1]

Options

• 10° C

• 20° C

• 5° C

• 40° C

Answer 1

10° C

Explanation:

c = `Q/("m"Delta"T")`

Q = `"cm"Delta"T"`

For copper,

Q = 420 × 50 × 10^-3× 10 = 210 calories

For water,

`Delta"T" = "Q"/"cm"`

`Delta"T" = 210/(4200 xx 5 xx 10^-3)` = 10° C