The temperature of the quartz object will increase by 232°C
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If the temperature of the lead object increases by 4.0°C , by how much does the temperature of the quartz object increase?
Given that :
ΔVl = ΔVq ( change in volume )
Δtl = 4°C
Final volumes after expansion
Vq = Vqi ( 1 + ∝q*Δtq )
Vl = Vli( 1 + ∝l*Δtl )
∴ ΔVq = Vqi*∝q*Δtq ----- ( 1 )
ΔVl = Vqi*∝l* Δtl ------ ( 2 )
Dividing equation 1 and 2
Δtq = ( ∝l * Δtl ) / ( ∝q ) -------- ( 3 )
where ; ∝l ( coefficient of volume expansion for lead ) = 87 * 10^-6 I / °C
∝q ( coefficient of volume expansion for quartz) = 1.5 * 10^-6 1 / °C
Δtl = 4°C
Insert values into equation ( 3 )
Δtq = 232°C
Hence we can conclude that the temperature of the quartz object will increase by 232°C .
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